Saturday
April 19, 2014

Search: calculate pH at equivalence point in titrating 0.120M solution of 0.08M HBr

Number of results: 69,662

chemistry
A certain acetic acid solution has pH = 2.14. Calculate the volume of 0.0900 M KOH required to reach the equivalence point in the titration of 35.0 mL of the acetic acid solution.
Thursday, April 3, 2014 at 8:17pm by Tessa

Chemistry
The problem is too long to work; however, if you will tell me what you know about it and what you don't understand about it I can help you through it. The process involves three stages. 1. at the beginning, 0 mL KOH added. 2. hydrolysis of the salts at each equivalence point. ...
Monday, March 11, 2013 at 9:26pm by DrBob222

Chemistry, pH, Buffers
Is the 18.47 mL base used the amount to titrate to the equivalence point. Is the pH at the equivalence point 3.49 or is that some other point on the titration curve? I don't get a molar mass of 165.3 from the data above. Is there more data? I suspect so. Please type the entire...
Monday, April 2, 2012 at 11:44am by DrBob222

chemistry
A volume of 100 of 1.00 solution is titrated with 1.00 solution. You added the following quantities of 1.00 to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.
Saturday, April 19, 2008 at 6:27pm by samantha

AP Chemistry
I would use the Henderson-Hasselbalch equation here. Have you had that in your course? pH = pKa + log[(base)/(acid)] Substitute pH = 5 (the middle point of where the indicator turns color) and calculate (base)/(acid). pKa for acetic acid (which I will call HAc) is 4.74. I ...
Saturday, April 3, 2010 at 7:07pm by DrBob222

chemisry
A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent. 100 ml of a 0.1M solution of the above acid is titrated with 0.1 M NaOH. After an ...
Sunday, February 12, 2012 at 3:16pm by ashley

Chemistry
Exactly 100 mL of 0.14 M nitrous acid (HNO2) are titrated with a 0.14 M NaOH solution. Calculate the pH for the following. (a) the initial solution (b) the point at which 80 mL of the base has been added (c) the equivalence point (d) the point at which 105 mL of the base has ...
Saturday, March 17, 2012 at 12:33am by Tanya

chemistry
Using H2CO3 as an example, the pH at the first equivalence point where H2CO3 ==>H^+ + HCO3^- H+ at equivalence point is sqrt(k1k2) or pH = 1/2(pk1+pk2). Do the same thing with H2A. Simply weigh out the NaHA salt (half neutralized. pH = 1/2(pK1 + pK2) = 6
Tuesday, July 27, 2010 at 6:18pm by DrBob222

college chemistry
Suppose you have a 0.200 M solution of the nitrogen-containing weak base NX. Suppose you wish to titrate 25 mL of this 0.200 M NX solution with a 0.100 M solution of the strong acid HNO3. Given that the Kb value of NX is 6.50x10-5, complete each of the following: a) What ...
Wednesday, April 21, 2010 at 5:58pm by Aubree

chemistry
a. A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent. b. 100 ml of a 0.1M solution of the above acid is titrated with 0.1 M NaOH. After an...
Sunday, February 12, 2012 at 4:29pm by billy

Chemistry
A student titrates 0.025 M KOH into 50.00 ml of a solution of unknown weak acid HA. Equivalence point is reached when 30.00 ml of the KOH has been added. After 15.00ml of the KOH has been added the pH of the mixture is 3.92. A. What is the concentration of HA ? B. What is Ka ...
Thursday, March 10, 2011 at 8:18am by Patrick

Chemistry
A student titrates 0.025 M KOH into 50.00 ml of a solution of unknown weak acid HA. Equivalence point is reached when 30.00 ml of the KOH has been added. After 15.00ml of the KOH has been added the pH of the mixture is 3.92. A. What is the concentration of HA ? B. What is Ka ...
Thursday, March 10, 2011 at 1:14pm by Patrick

chemistry
For the first equivalence point, (H^+) = sqrt(k1k2) For the second equivalence point, pH is determined by the hydrolysis of the salt, in this case Na2M M^-2 + HOH ==> HM^- + OH^- Write the Keq (Kb) expression, set up an ICE chart, solve for OH^-, convert to pOH, then to pH.
Sunday, March 8, 2009 at 1:40am by DrBob222

chemistry
11.During an acid-base titration, 25 mL of NaOH 0.2 M were required to neutralize 20 mL of HCl. Calculate the pH of the solution for each of the following: 12.Before the titration. 13.After adding 24.9 mL of NaOH. 14.At the equivalence point. 15.After adding 25.1 mL of NaOH. ...
Wednesday, May 8, 2013 at 9:18pm by Tina

College Chemistry
Assuming that Ka is 1.85 *10-5 for acetic acid, calculate the pH at one-half the equivalence point and at the equivalence point for titration of 50mL of 0.100M acetic acid with 0.100M NaOH.
Sunday, October 14, 2012 at 9:29am by Kenetria

Chemistry
We can't draw pictures/graphs on the board. Here is how the titration of a weak acid with a strong base looks. Scroll down about half-way. http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html The secret to calculating the pH at selected points along the curve comes ...
Tuesday, April 19, 2011 at 12:28am by DrBob222

CHEM 136
A sample of 25.00 mL of 0.100 M HNO2(in a flask) is titrated with 0.150 M of NaOH solution at 25 degrees. 1) calculate the volume(Ve) of the NaOH solution needed to completely neutralize the acid in the flask. 2)calculate the pH for (a)the initial acid solution in the flask (b...
Tuesday, November 23, 2010 at 4:49pm by Jefferson

Chemistry
0.50 M HI with 0.10 M KOH pH at equivalence point? I had another problem like this and I tried to do it the same way b/c I thought that was the right thing to do but it seems as if its not? Can you help? HI + KOH ==> KI + H2O. At the equivalence point we will have KI (the ...
Monday, April 2, 2007 at 2:43pm by Paul

Chemistry
Calculate the theoretical pH at the starting point and at each equivalence point for phosphoric acid, acetic acid, and sulfuric acid. Please show the equilibrium equation, the Ka, and the pH calculation.
Monday, December 12, 2011 at 7:26pm by Shana

Chemistry (College) URGENT
1.6g of an unknown monoprotic acid (HA) required 5.79 mL of a 0.35 M NaOH solution to reach the equivalence point. Determine the molar mass of the acid and given that the pH at the half way point to the equivalent point is 3.86. Calculate the Ka of the unknown acid.
Monday, October 29, 2012 at 6:45pm by Tiffany

AP Chemsitry
A solution of an unknown monoprotic weak acid was titrated with 0.100 M NaOH. The equivalence point was reached when 37.48 ML of base had been added. From a second buret, exactly 18.74 of 0.100 M HCl were added to the titration solution. The pH was then measured with a pH ...
Sunday, April 1, 2012 at 5:15pm by Keith

college
Both of these problems have moles acid = moles base which means you are at the equivalence point of each. Therefore, the pH is determined by the pH of the salt. Write the hydrolysis equation, set up an ICE chart, and solve for (H^+), then convert to pH. One important point to ...
Monday, June 21, 2010 at 9:07am by DrBob222

Chemistry
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown ...
Sunday, February 24, 2013 at 8:42pm by Kat

Chemistry
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown ...
Sunday, February 24, 2013 at 9:19pm by Summer

chemistry
You have 5 points on the titration curve. You must recognize where you are on the curve. At zero mL, you have pure Ba(OH)2 so the pH will be determined by that concn. At the other points, calculate how moles Ba(OH)2 you have initially, the moles HCl added. Calculate moles Ba(...
Monday, March 28, 2011 at 1:19pm by DrBob222

Chemistry
The equation is C6H5NH2 + HCl ==> C6H5NH3^+ + Cl^- So at the equivalence point, the solution is C6H5NH3^+ (the amine salt). What is its concentration. It will be You don't specify how much of the aniline is being titrated BUT since it is 0.2 M and the HCl is the same, then ...
Thursday, October 15, 2009 at 3:35pm by DrBob222

chemistry
A 30 mL sample of .165 M propanoic acid is titrated with .300M KOH. Calculate the pH at each volume of added base: 0 mL, 5 mL, 10, equivalence point, one-half equivalence point, 20 mL, 25 mL. Use calculations to make a sketch of the titration curve.
Wednesday, October 23, 2013 at 11:03pm by Anonymous

chem
According to my table, bromocresol green changes pH between 3.8 and 5.4. You can calculate the approximate pH for the equivalence point for CH3COOH vs NaOH and that is about 8.5 or so. What do you think?
Tuesday, November 23, 2010 at 10:13pm by DrBob222

Chemistry
..........HNO2 + H2O ==> H3O^+ + NO2^- initial...0.150M..........0........0 change.....-x.............x........x equil...0.150-x...........x.........x Ka = (H3O^+)(NO2^-)/(HNO2) Substitute from the ICE chart and solve for x = (H3O^+). Then pH = -log(H3O^+) 3. Calculate ...
Friday, April 6, 2012 at 4:57pm by DrBob222

AP Chemistry
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown ...
Sunday, February 24, 2013 at 2:42pm by Sarah

college chemistry
In class, we discussed the titration of a 50.0 mL sample of 0.100 M HCL with a 0.100 M solution of NaOH. How would this system change if we used a 0.100 M solution of Sr(OH)2 instead of NaOH? To answer this question, complete each of the following: a) What volume of 0.100 M Sr...
Wednesday, April 21, 2010 at 5:59pm by Aubree

chemistry
For any titration you want a sharp change in pH at the equivalence point; therefore, you KNOW #3 can't be right. I'm not sure what #1 means but I would assume sloping upward would mean a more or less gradual change from beginning to end; therefore, I wouldn't pick #1. The ...
Wednesday, June 11, 2008 at 10:28pm by DrBob222

Chemistry
Can you calculate volume to get to the equivalence point? Do that first. a. First stage: at the beginning. .........HA ==> H^+ + A^- initial..0.025..0.....0 change...-x.....x.....x. equil...0.025-x..x....x Substitute into the Ka expression and solve for x, then convert to ...
Wednesday, April 11, 2012 at 7:03pm by DrBob222

Chemisty
In titrations of acids and bases, what is the difference between the end point of the titration, and the equivalance point? Also, what is chemically occuring during the buffer zone? Thanks! Good question and one that students sometimes have trouble with. The equivalence point ...
Tuesday, May 15, 2007 at 9:47pm by Jessie

chemistry
A 1.0 M acetic acid solution (CH3COOH, pKa = 4.7) is neutralized by dissolving NaOH(s) (a strong base) in the solution. Estimate the pH at the equivalence point of the neutralization process: 4.7 7.0 9.3 Which one of the following statements best explains your prediction? 1. ...
Monday, March 31, 2014 at 4:23pm by bekah

Chemistry
The number of moles NaOH needed to reach the equivalence point is mol, which means we must add liters of NaOH. The total volume of solution at the equivalence point will be liters. From the above information, the molarity of NaF at the equivalence point is M.
Wednesday, March 13, 2013 at 11:48pm by Elizabeth

Chemistry
This is a strong acid/strong base titration so the pH at the equivalence point will be 7.00. The first thing to do is to calculate where the equivalence point is.The following will do that. mL acid x M acid = mL base x M base mL acid x 0.125 = 30.00 x 0.150 mL acid = ? Then ...
Thursday, November 17, 2011 at 4:12pm by DrBob222

Chemistry
In a experiment to determine the molecular weight and the Ka for ascorbic acid (vit. c.) a student dissolved 1.3713g of the monoprotic acid in water to make 50 mL of solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 mL of the...
Saturday, May 19, 2007 at 6:57pm by Tri

Chemistry
There is a system and sets of formulas depending upon the problem. 0.3M HCl. HCl is a strong acid, meaning it ionizes 100%, which means H^+ = 0.3 and Cl^- is 0.1M So pH = -log(H^+). Plug in H^+ and solve. For the equivalence point, you must recognize what you have at the ...
Monday, March 18, 2013 at 8:27pm by DrBob222

Chemistry
A 30.00ml sample of 0.1234 M hypobromous acid (HBrO) is titrated with a 0.2555 M KOH solution. The Ka for HBrO is 2.5*10^-9 Calculate the pH of teh titration mixture at the equivalence point.
Saturday, April 17, 2010 at 11:28am by Anonymous

Chemistry
q1. The label on an antacid remedy states that each tablet contains 750 mg of aluminum hydroxide. Calculate the volume of stomach acid, 0.10 mol/L HCl(aq), which can be nuetralized by one antacid tablet. Q2. Compare equivalence point to endpoint q1. The label on an antacid ...
Friday, September 22, 2006 at 5:06pm by Brittany Rusk

chemistry
ph at equivalence point when 25 ml of a 0.175M solution of acetic acid is titrated with 0.10M of NaOH at its end point
Monday, March 19, 2012 at 3:28pm by Sharnam

College Chemistry
Consider the titration of 29.00 mL of aqueous ammonia with 18.70 mL of 0.1250 M HCl (aq). Kb for NH3 (aq) is 1.8 x 10^-5. A.) What is the formula of the solute at the equivalence point? B.) What is the balanced equation for the ionization of the cation of the solute? C.) What ...
Tuesday, April 13, 2010 at 6:39pm by MiShell

Chem 20
First, we can't draw plots on the board. I don't know what you mean by a "concn of a titration curve." The pH at the equivalence point of HNO3 and KOH titration is 7.00 and the equivalence point occurs at mLHNO3 x MHNO3 = mLKOH x M KOH The concn of the KNO3 at the equivalence ...
Friday, January 7, 2011 at 1:03am by DrBob222

Chemistry
What is the equivalence point in an acid-base titration? I'm supposed to do a virtual lab online and it tells you to find the equivalence point, but I don't know if that means when the pH is 7 or when the indicator changes colors.
Tuesday, March 24, 2009 at 10:05pm by Katie

Chem Titration and pH
Calculate the pH at 0mL, 5mL,...40mL for a 10.0mL aliquote of 0.100M Na3AsO4 (weak base) titrated with 0.100M HCl pKa1 = 2.25 pKa2 = 6.77 pKa3 = 11.60 Thank you First you should write equations to know where we are in the titration. At the beginning, we have the hydrolysis of ...
Saturday, October 7, 2006 at 6:15pm by julie

Chem
You are titrating an unknown weak acid you hope to identify. Your titrant is a 0.0935 mol/L NaOH solution, and the titration requires 22.3 mL to reach the equivalence point. How many moles of acid were in your sample?
Sunday, October 3, 2010 at 12:19am by Kim

chemistry
You are titrating an unknown weak acid you hope to identify. Your titrant is a 0.0935 mol/L NaOH solution, and the titration requires 39.9 mL to reach the equivalence point. How many moles of acid were in your sample?
Tuesday, October 2, 2012 at 6:11pm by lucy

Chemistry
What is the pH for the potassium propionate solution at the equivalence point?
Tuesday, May 3, 2011 at 4:16pm by Mira

chemistry
If you over shot the end point in which you are titrating HCl with NaOH, there is no H^+ to neutralize at the equivalence point of the titration. So some extra Mg(OH)2 COULD form but if you over shot on purpose, why would that cause a problem. If I over shot the end point, I ...
Monday, October 18, 2010 at 11:23pm by DrBob222

Chemistry
One problem I am having in helping you is that I can't figures how you started with 20 mL of 0.1M H3PO4 and it has a pH of 4. The pH of a 0.1 M H3PO4 soln is <2. The second problem is that you give a pk1 but not a pk2 (suggesting pk2 is not needed) BUT the pH of H3PO4 in ...
Thursday, September 23, 2010 at 3:21pm by DrBob222

chemistry
Glutamate (Glu–) is the conjugate base form of glutamic acid (HGlu). The Ka of glutamic acid is 5.012 x 10^–5. You titrate 50 mL of 0.10 M sodium glutamate solution with 0.05 M HCl solution. Calculate the pH of the solution at the equivalence point. Ive been trying to figure ...
Saturday, July 28, 2012 at 11:34am by kellie

Analy Chem
Find mols NaOH, subtract mols HBr. If mols NaOH are greater then pOH = -log(OH^-) and convert to pH by pH + pOH = pKw = 14. If mols HBr are in excess, pH = -log(H^+). Those three steps will do all of the points except for the equivalence point. That one is done by the ...
Wednesday, April 11, 2012 at 6:54pm by DrBob222

chemistry
millimoles HNO3 = mL x M = 50.00 x 0.3M = 15 millimoles. mmoles KOH = mL x 0.6M = mL = 0, 0mmoles KOH mL = 15.5, 9.3 mmoles mL = 25.0, 15 mmoles mL = 40.0, 24 mmoles I'll do the 15.5 or 9.3 mmoles KOH added. ..........HNO3 + KOH ==> KNO3 + H2O initial...15......0........0...
Friday, December 2, 2011 at 6:50pm by DrBob222

Analytical chemistry help!
Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN (Ka of HCN = 6.2 x 10-10) with a) 4.20 mL of 0.438 M HClO4 and b) 11.82 mL of 0.438 M HClO4. What is the pH at the equivalence point with 0.438 M HClO4?
Saturday, November 26, 2011 at 4:37pm by Remy

Chemistry
I am doing similar homeowrk and I was wondering how you calculate the ph after the equivalence point?
Thursday, November 17, 2011 at 4:12pm by hannah

Chemistry
I can't draw on this forum how these things look; however, remember that with the HA, adding NaOH produces a buffered solution. Because it's a weak acid the initial pH (before any NaOH is added) is higher and the addition of NaOH forms NaA. The mixture of unreacted HA and the ...
Thursday, March 3, 2011 at 11:10pm by DrBob222

Chemistry
While titrating 25.00mL of a weak acid, HA, with 0.1500M NaOH, you reach equivalence point after adding 27.00mL of the NaOH. The pH of the acid initially was 2.48. What is the dissociation constant of the acid?
Monday, October 27, 2008 at 9:38pm by Sam

Chemistry
The secret to doing this is to recognize what you have at each titration point. If we let HB stand for butanoic acid, then the equation is HB + NaOH ==> NaB + H2O a. at the beginning you have just HB. Set up an ICE chart ............HB ==> H^+ + B^- initial....0.1..........
Wednesday, November 23, 2011 at 12:50am by DrBob222

chemistryyy
HBrO + KOH ==> KBrO + H2O 1)What do you have at the equivalence point? That is a soln of KBrO. It is the salt of a strong base and a weak acid; therefore, the solution MUST be basic. 2) The pH at the equivalence point is determined by the hydrolysis of the salt...
Wednesday, September 21, 2011 at 2:59pm by DrBob222

Chemistry!!
At the equivalence point we have the salt, CH3NH3Cl and it will hydrolyze to give you an acidic solution. (salt) at the equivalence point is 0.09M. .........CH3NH3^+ + H2O == H3O^+ + CH3NH2 I.........0.09..............0........0 C...........-x..............x........x E.......0...
Friday, May 11, 2012 at 8:32pm by DrBob222

Chemistry
For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaoH, calculate (a) the initial pH; (b) the pH when neutralization is 50% complete; (c) the pH when neutralization is 100% complete; and (d) the pH when 1.00 mL of NaOH is added beyond the equivalence point.
Tuesday, August 11, 2009 at 7:25am by ChemHurts

152
The pH of HClO at the beginning of the titration is calculated from HClO as a single weak acid. Additions of NaOH form NaClO, which is a weak acid and its salt, therefore, the Henderson-Hasselbalch equation is used. At the equivalence point, the acid is completely neutralized ...
Wednesday, March 31, 2010 at 7:36pm by DrBob222

Balthazar
1. Do you mean 0.150 m or 0.150M? You need to get your cap keys in order. I think you MUST mean M and I will work it that way. (By the way, M stands for molarity; m stands for molality.) a. Write the equation and balance it. b. Calculate mL NaOH to reach the equivalence point...
Thursday, August 1, 2013 at 2:09pm by DrBob222

Chemistry 2
A titration is performed by adding 0.124 M KOH to 40 mL of 0.159 M HNO3. a) Calculate the pH before addition of any KOH. b) Calculate the pH after the addition of 10.26, 25.65 and 50.29 mL of the base.(Show your work in detail for one of the volumes.) c) Calculate the volume ...
Tuesday, April 2, 2013 at 10:44pm by Antonio

science
1. Which of the following is TRUE regarding this situation: Solution A has a pH of 7.38, and Solution B has a pH of 7.42? a.Solution B is more acidic than Solution A b.The pH of Solution A falls within the homeostatic pH range for extracellular body fluids, but the pH of ...
Monday, January 28, 2013 at 12:21am by a

chemistry college
The secret to these is to recognize what you have in the solution at each addition. The first thing you do is to determine where the equivalence point is; i.e., number of mL HCl added to get to the equivalence point. 20 mL--you have a mixture of NH3 and NH4Cl. That makes a ...
Wednesday, December 4, 2013 at 5:33pm by DrBob222

Chemistry
The equivalence point of a titration occurs when the moles of one component you started with equals the moles of the titrant. This point may or may not be at pH=7.0. The END POINT occurs when the indicator changes color, where ever that may be. We try to match the end point ...
Tuesday, March 24, 2009 at 10:05pm by DrBob222

Chemistry
Regular-strength chlorine bleach is a 5.25% solution of NaClO. how many milliliters of 0.02005 M sodium thiosulfate will be required to reach the equivalence point in the titration of a 1.000ml sample of Bleach by reacting the ClO- ions with I, then titrating the resulting I2 ...
Monday, October 10, 2011 at 2:26am by Jen

chemistry
) At the equivalence point in the titration of a strong acid with a strong base, do you expect the pH to be acidic, basic, neutral? Explain your answer. If the titration involves a strong acid with a strong base, the pH = 7 at the equivalence point. The pH will be neutral ...
Tuesday, March 3, 2009 at 4:36pm by Anonymous

chemistry
An HNO3 solution has a pH of 3.04. What volume of 0.015 M LiOH will be required to titrate 89.0 mL of the HNO3 solution to reach the equivalence point?
Thursday, January 5, 2012 at 8:11pm by Anonymous

chem
The procedure for finding pH at the beginning is correct but you made a math error somewhere. It appears you may have used Kb and not Ka. 1.5 x 10^-5 = (H^+)^2/0.105 and solve for (H^+). The pH is between 2 and 3. Volume to reach the equivalence point is mL x M = mL x M for ...
Monday, March 1, 2010 at 11:38pm by DrBob222

Chemistry
a)HCl --> H^+ + Cl^- HCl is 100% ionized; therefore, (HCl) = (H^+) = 0.8M pH = -log(H^+) = ?? b)to start. moles HCl = M x L = 0.800*0.040 = 0.032. moles KOH added in 5 mL = M x L = 0.600*0.005 = 0.003 ................HCl + KOH ==> KCl + H2O initial......0.032....0...
Friday, April 1, 2011 at 1:18pm by DrBob222

chemistry
A 20.0 mL sample of an aqueous HClO3 solution is titrated with a 0.0275 M KOH solution. The equivalence point is reached with 28.5 mL of the base. The pH of the HClO3 solution, before titration, is
Tuesday, March 13, 2012 at 5:15pm by Maggie

chemistry
A. Strong Base 1.) What is the concentration of a solution of KOH for which the pH is 11.89? 2.) What is the pH of a 0.011M solution of Ca(OH)2? B. Weak Acid 1.) The pH of a 0.060M weak monoprotic acid HA is 3.44. Calculate the Ka of the acid. 2.) The pH of 0.100M solution of ...
Thursday, January 20, 2011 at 3:37am by jaycab

chemistry
A solution of NaOH with pH 13.68 requires 35.00 mL of 0.128 M HClO_4_ to reach the equivalence point. What is the volume of the NaOH solution?
Thursday, March 8, 2012 at 10:40pm by Savannah

Chemistry
How do I do this?? Calculate the pH at the equivalence point for the titration of 0.20 M aniline (C6H5NH2, Kb = 3.8 x 10-10) with 0.20 M HCl. [Hint: remember the dilution factor caused by addition of titrant to the aniline solution.]
Thursday, October 15, 2009 at 3:35pm by Jamie

Chemistry
If we call methylamine, MeNH2, then the pH at the equivalence point is determined by the hydrolysis of the salt, MeNH3Cl formed at the equivalence point. This is Ka for MeNH3Cl .........MeNH3^+ + H2O -->MeNH2+ + H3O^+ initial..0.06M..............0........0 change.....-x...
Sunday, November 13, 2011 at 10:10pm by DrBob222

chemistry
First I would determine mL for the equivalence point. Use the Henderson-Hasselbalch equation for all point up to the equivalence point. Use the hydrolysis of (Kb) NaClO for the equivalence point. Use the excess NaOH for anything past the equivalence point; however, I don't ...
Thursday, March 10, 2011 at 10:31pm by DrBob222

Chem
Theoretically, we want the end point and the equivalence point to be the same but if the titration curve is steep enough at the equivalence point, we can get away with using an indicator that produces an end point very close to the equivalence point (in terms of extra titrant ...
Saturday, May 15, 2010 at 3:06pm by DrBob222

Chemistry
Sodium hypochlorite, NaClO, is added as a disinfectant to the water supply. You are analyzing a sample from the supplier to verify its purity and you prepare to titrate 250.0 mL of a solution containing 1.86 g of NaClO with a 1.00 M HCl solution. What is the pH of the solution...
Monday, February 13, 2012 at 2:50am by Rosa

Chem
KOH + HBr ==> H2O + NaBr. 50 x 0.1 = 50 x 0.1 so this is at the exact equivalence point which means that the pH is determined by the hydrolysis of the NaBr salt. Neither Na^+ nor Br^- is hydolyzed in H2O solution; therefore, the pH is not changed from neutral H2O so pH = 7.0
Monday, December 3, 2012 at 7:23am by DrBob222

Chem
if 0.24o moles of a monoprotic weak acid is titrated with NaOH, what is the PH of the solution at the 1/2 equivalence point.
Monday, March 10, 2014 at 7:04pm by Joe

chemistry
mols NaOH = M x L mols CH3COOH = mols NaOH. mols CH3COOH = M x L You took 25 mL, you measure the mols CH3COOH, calculate M CH3COOH. I presume you are to graph the results. I don't know what instructions you have been given but you must find the equivalence oint for the ...
Tuesday, April 8, 2008 at 11:39pm by DrBob222

Chemistry
CH3COOH + NaOH ==> CH3COONa + H2O So the pH at the equivalence point will be determined by the hydrolysis of the salt. CH3COO^- + HOH ==> CH3COOH + OH^- Kb = (KwKa) = (OH^-)(CH3COOH)/(CH3COO^-) Set up an ICE chart, substitute into the Kb expression, and solve for x, then...
Wednesday, December 15, 2010 at 5:02pm by DrBob222

chemistry
They aren't always different but, by design, we try to put one as close to the other as we can. The equivalence point is the theoretical point at which the two substances EXACTLY react with no excess of either reagent. Since a large number of solutions react with no visible ...
Tuesday, April 6, 2010 at 5:41pm by DrBob222

Chemistry
What is the pH at the equivalence point when 0.10 M HNO_3 is used to titrate a volume of solution containing 0.30 g of KOH?
Wednesday, November 21, 2012 at 8:02pm by Josh G.

Chemistry
Determine where the equivalence point is. Note the volume. mmoles OH^- added in excess is 5.0 mL x M = ?? total volume = mL to arrive at the equivalence point + 5.0 from the base mmoles/total mL = M of OH^- pOH = -log(OH^-) and pH + pOH = pKw = 14. Solve for pH.
Sunday, February 27, 2011 at 6:21pm by DrBob222

chemistry
On the axis below, draw the titration curve for the titration of CH3COOH (a weak acid) with NaOH (a strong base). In the titration, the equivalence point occurs when 40 mL of NaOH has been added. Make sure to label your x- and y-axis and put numbers on your grid. Make sure ...
Monday, April 23, 2012 at 12:36am by Anonymous

Chemistry
Any combination of acid and base titration will give a SHARP end point caused by a large increase (or decrease depending upon what is titrating what) of about 4 or 5 pH units EXCEPT for the titration of a weak acid with a weak base. With the WA/WB titration it is essentially ...
Thursday, May 12, 2011 at 7:28pm by DrBob222

Chemistry
calculate the pH at 0ml at the quivalence point and at 40ml in a titration of 25 ml of .120M formic acid with .105M NaOH (kA of formic acid= 1.8x10^-4)
Sunday, March 1, 2009 at 1:34pm by Gagan

Chemistry
You could titrate a sample of vinegar with a known standard of NaOH. You can look up the details of a titration. The pH at the equivalence point is determined by the hydrolysis of the salt, in this case, acetate (C2H3O2^-). ......Ac^- + HOH ==> HAc + OH^- Kb for Ac^- (Kw/Ka...
Sunday, August 4, 2013 at 8:07pm by DrBob222

sceince 2
6. When aqueous solutions of table salt and silver nitrate are mixed, a white solid forms. This solid is ____. (1 point)soluble a precipitate an alloy a solute 7. Stainless steel is an example of a ____ solution. (1 point)liquid-solid solid-liquid solid-solid gas-solid 8. ...
Tuesday, April 17, 2012 at 1:34pm by NIKE

chemistry
HCl + NaOH ==> NaCl + H2O At zero mL. YOu have 0.1 M HCl so the pH is pH = -log(H^+). Since HCl is a strong acid, H^+ = 0.1 M and pH = 1. All of the others are done this way: a. moles HCl initially = M x L = 0.025 x 0.1 = 0.0025 moles HCl. b. moles NaOH added. (at 5 mL this...
Friday, April 30, 2010 at 12:08pm by DrBob222

CHEM
Are you sure that's 0.900 M NaOH? and not 0.0900 M? I'll assume 0.900 is correct. And I'll write acetic acid, HC2H3O2, (or CH3COOH) as HAc. Millimols HAc initially = 25 x 0.1 = 2.5. millimols NaOH added = mL x 0.900 The first thing to do is to determine the equivalence point. ...
Monday, October 1, 2012 at 12:54pm by DrBob222

Chemistry help!!
Surely you mean Ve = volume at the equivalence point. The Henderson-Hasselbalch equation is pH = pKa + log(base)/(acid) If pH = pKa, then log B/A = 0 and B/A = 1 If pH = pKa + 1, then log B/A = 1 and B/A = 10 If pH = pKa -1, then log B/A = -1 and B/A = 0.1 ..........HA + OH...
Tuesday, November 29, 2011 at 11:01pm by DrBob222

Chemistry
Consider a solution that contains both C6H5NH2 and C6H5NH3+. Calculate the ratio [C6H5NH2]/[C6H5NH3+] if the solution has the following pH values. (Assume that the solution is at 25°C.) (a) pH = 4.70 (b) pH = 5.26 (c) pH = 5.42 (d) pH = 4.96
Wednesday, March 13, 2013 at 2:47am by Margaret

Chemistry
To be honest with you, I've never heard the term equivalence volume used. However, my best educated guess is that the equivalence volume is the volume of a titrate required to reach the equivalence point. By the way, and this is a little picky, when you titrate with an ...
Saturday, February 16, 2008 at 6:59pm by DrBob222

Chemistry lab
1) Using the 1st equivalence point from part a calculate the molarity of the phosphoric acid. Molarity of NaOH = .1523 M Volume of phosphoric acid used 10.0mL Volume at equivalence point 7.410 mL and 14.861 mL 2) Using the 2nd equivalence point from part a, calculate the ...
Wednesday, April 11, 2012 at 11:16am by Maura

Chemistry
Calculate the pH at the equivalence point for the titration of 0.130M methalamine (CH3NH2) with 0.13M Hcl. The Kb of methyl amine is 5.0x10-4.
Monday, March 18, 2013 at 4:44pm by Joe

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