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

Number of results: 46,478

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

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 solution of NaOH with pH 13.68 requires 35.00 mL of 0.128 M HClO_4_ to reach the equivalence point. How do you calculate [Na^+^] and [ClO_4_^-^] at the aquicalence point? (volumes are additive)
Thursday, March 8, 2012 at 11:50pm by Savannah

chemistry
0.5 L of a 0.30 M HCl solution is titrated with a solution of 0.6 M KOH. a)What is the pH before addition of KOH? b)What is the total number of moles of acid? c)What is the pH after addition of 125 mL of KOH solution? d)What volume of the KOH solution is required to reach the ...
Thursday, April 28, 2011 at 1:05pm by Joe

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

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

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
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
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

Chemistry
The key to this type problem is to recognize what is in the solution at the point of calculation. 0 mL must be pure formic acid. The equivalence point is the salt, sodium formate. so the pH is determined by the hydrolsis (the Kb) of the salt. At 40.00 mL, (past the equivalence...
Sunday, March 1, 2009 at 1:34pm 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...
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 ...
Tuesday, April 6, 2010 at 5:41pm by DrBob222

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
The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.460 moles of a monoprotic weak acid (Ka = 3.8 × 10-5) is titrated with NaOH, what is ...
Wednesday, April 10, 2013 at 11:55am by sara

Chemistry
The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.220 moles of a monoprotic weak acid (Ka = 5.7 × 10-5) is titrated with NaOH, what is ...
Monday, April 8, 2013 at 3:41pm by Felicia

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
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
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
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 ...
Friday, April 30, 2010 at 12:08pm by DrBob222

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 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 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
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 II
Calculate the pH at the equivalence point in the titration of 50ml of 0.20 M methylamine (Kb=4.3*10^-4) with a 0.40 M HCL solution.
Friday, May 3, 2013 at 4:24pm by Katie

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

Chemistry
To do zero mL base. HCOOH ==> H^+ + HCOO^- Prepare an ICE chart and substitute into the Ka expression. Ka = (H^+)(HCOO^-)/(HCOOH) Solve for (H^+) and convert to pH. For all of the others. The equation is HCOOH + NaOH ==> HCOONa + H2O 1. First determine where the ...
Sunday, April 3, 2011 at 5:13pm by DrBob222

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 II
at the beginning you have a "pure" solution of pyridine with Kb = ? ........BN + HOH ==> BNH^+ + OH^- I....0.370M...........0......0 C.......-x............x......x E....0.370-x..........x......x Kb = etc. Substitute into the Kb expression and solve for x = (OH...
Thursday, May 2, 2013 at 10:54pm by DrBob222

chemistry
Ca(OH)2 ==> Ca^2+ + 2OH^- Ksp = (Ca^2+)(OH^-)^2 Assuming you titrated with HCl, the equation is Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O So at the equivalence point the pH = 7.45. I would convert to pOH (pH + pOH = pKw = 14), then pOH = -log(OH^-) to obtain (OH^-). For Ca...
Tuesday, November 1, 2011 at 10:07pm by DrBob222

chemistry
Did you add a particular volume of titrant and measure the pH of the resulting solution? If not, how did you perform the experiment. The usual way of determining the Ka is to look up the pH at the halfway point to the equivalence point.
Friday, May 9, 2008 at 9:47pm by DrBob222

chemistry
CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH of the solution to be 11.77. (a) Write ...
Thursday, April 4, 2013 at 6:08pm by ashdawg

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

AP Chemistry
CH3NH2(aq)+H2O(l)=>CH3NH3+(aq)+OH-(aq) Kb=4.4 x 10^-4 Methylamine, CH3NH2, is a weak base that reacts with water according to the equation above. A student obtains a 50.0 mL sample of a methylamine solution and determines the pH of the solution to be 11.77. (a) Write ...
Thursday, May 2, 2013 at 12:41am by Christina

Titration
This is the titration of a strong base with a strong acid. Begin: 100 mL x 0.100M NaOH = 10 millimoles. millimoles HBr added = 1.00M x mL = ? .............NaOH + HBr ==> NaBr + H2O initial......10mmol...0.......0......0 added.................0 change........-0...
Monday, January 23, 2012 at 3:46pm by DrBob222

chemistry
For the titration of 20.0 mL of 0.1500 mol/L NH3(aq) with 0.1500 mol/L HI(aq)(the titrant), calculate a) the pH before any HI(aq) is added b) the pH at the equivalence point
Sunday, April 6, 2008 at 2:08am by Ross

Chemistry
Equivalence point Volume of 14mL and a conductivity of 44. Sulfuric Acid=0.020M and 13mL Barium Hydroxide=100mL -Determining the volume of Sulfuric acid added to the equivalence point. -Calculating moles of sulfuric acid added at the equivalence point. -Calculating the moles ...
Wednesday, January 5, 2011 at 9:57pm by Anonymous

Chemistry
Here is the way you know. acid + base ==> salt + water. If the acid is strong and the base is weak, the salt will be acidic and the equivalence point will be <7. If the base is strong and the acid is weak, the salt will be basic and the equivalence point will be...
Sunday, June 13, 2010 at 1:53pm by DrBob222

Chemistry
Calculate the pH at the equivalence point for the titration of 0.120 M methylamine with 0.120 M HCl (kb of methyalamine is 5.0 x 10^-4)
Sunday, November 13, 2011 at 10:10pm by Mariana

Chemistry
A solution of acetic acid having a concentration of about 0.2M is to be titrated using 0.200M NaOH. Select an indicator for the titration. what salt will the solution contain at the equivalence point? what is the approximate concentration of this salt at the equivalence point...
Saturday, May 15, 2010 at 2:00pm by Alison

Chemistry
I don't think you can use the HH equation for the equivalence point. You have the base (acetate concn) but there is no acid and a number/0 is undefined. CH3COOH + NaOH ==> CH3COONa + H2O What you do is the pH at the equivalence point is due to the hydrolysis of th ...
Thursday, October 28, 2010 at 4:20pm by DrBob222

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

Chemistry
How do you figure equivalence volume? Do I add up all of the additions during my lab up to the point I am measuring (color change) to get the equivalence volume? I guess I am confused when it says calculate the equivalence volume, maybe it is very simply.
Saturday, February 16, 2008 at 6:59pm by Thomas

Chemistry
A complicated work but do-able. NaOH + HCl ==> NaCl + H2O. I think one must decide where we are at the equivalence point. Since the M NaOH is twice that of the HCl, that means the HCl will take twice as much to neutralize the NaOH. So that 35 mL is made up of 2 mL HCl ...
Sunday, May 2, 2010 at 11:52pm by DrBob222

chemistry
First determine L NaOH for the equivalence point. mols excess NaOH = M NaOH x L NaOH. New M NaOH = mols excess/total volume. (Note: total volume = volume to reach the equivalence point + 5 mL) That will give you the M of OH^-, then pOH = -log(OH^-) followed by pH + pOH = pKw...
Wednesday, March 21, 2012 at 7:35pm by DrBob222

Chemistry
Pyridine we will call Py When we titrate this base, the equivalence point will occur when only the salt is present. It will hydrolyze. You can calculate the pH at that point. PyH^+ + H2O ==> H3O^+ + Py Ka for PyH^+ = (Kw/Kb) where Kw is 1 x 10^-14 and Kb is the ...
Sunday, May 2, 2010 at 10:28pm by DrBob222

Chemistry
The details are missing but if you titrated with a curve, then you know where the equivalence point is. Read mL acid at that point, take 1/2 of that volume in mL, then read from the graph the pH at the half-way point. The pH = pKb at the half-way point.
Thursday, March 28, 2013 at 8:56pm by DrBob222

chemisty
3.14M NaN3 (pKb=9.28) is titrated with .663M HClO4 until it reaches the equivalence point. How do you calculate the pOH at the equivalence point. I tried and got something around 9 but its not correct i used a table but without liters given i have not idea where to go or what...
Saturday, March 16, 2013 at 11:32pm by sreedevi

chemistry
Q.1 What will be the pH at the equivalence point during the titration of a 100 ml 0.2M solution of CH3COONa with 0.2M of solution of HCl?(Ka = 2*10^-5) Q.2 Aniline behaves as a weak base.When 0.1M,50ml solution of aniline was mixed with 0.1M,25ml solution of HCl the pH of ...
Saturday, March 27, 2010 at 6:47am by gaurav

Chemistry
1. Yes. 2. Use the indicator that changes in the rage of pH that is the pH at the equivalence point of a titration/reaction. For example, Na2CO3 + HCl ==> NaHCO3 + NaCl The equivalence point is about 8.3 for th addition of ONE H so you would use phenolphthalein for this...
Sunday, April 5, 2009 at 6:19am by DrBob222

Chemistry III
Assume you dissolve 0.240 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.16 102 mL of solution and then titrate the solution with 0.153 M NaOH. C6H5CO2H(aq) + OH -(aq) C6H5CO2-(aq) + H2O(l) (a) What was the pH of the original benzoic acid solution? (b) ...
Tuesday, March 27, 2012 at 11:09pm by Kelly

chemisty
The salt present at the equivalence point is NH4Cl and there are 0.05 L x 0.15 M. That hydrolyzes at the equivalence point and that determines the pH. NH4^+ + HOH ==> NH3^+ + H3O^+ Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+) Plub in (NH4^+) from mols NH4Cl/liters (don't ...
Friday, December 7, 2007 at 6:33pm by DrBob222

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
25.0 ml of a 0.100 M solution of the weak acid, CH3OOH is titrated with a 0.10 M solution of the strong base, NaOH. Calculate the pH of the equivalence point.
Sunday, February 26, 2012 at 10:44am by ASood

chemistry
False. The end point is the point in the titration where the indicator "indicates" that the equivalence point has been reached. That may or may not be the real equivalence point.
Monday, December 5, 2011 at 11:45am by DrBob222

Chemistry
I don't think you have typed the entire question. What's missing is the volume of titrant needed to completely titrate (to the end equivalence point) the acid. Ka = (H^+) x (A^-)/(HA) You obtain (H^+ from pH in the problem. fraction A^- = 23.55/V fraction HA = 1- ...
Monday, April 29, 2013 at 5:00am by DrBob222

chemistry
a 205 mg sample of diprotic acid is dissolved in enough water to make 250 ml of solution. The pH of this solution is 2.15. A saturated solution of calcium hyrdoxide (Ksp=1.3*10^-6) is prepared by adding excess calcium hydroxide to water and then removing the undissovled solid ...
Sunday, March 4, 2012 at 8:09pm by cat

chemistry
"Taking 1/2 the volume at the equivalence point and reading the pH at that point will give you the pKa of the acid." Do you mean reading the pH at that point will give me the Ka* of the acid?
Tuesday, April 8, 2008 at 11:39pm by Anonymous

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
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 ...
Thursday, March 3, 2011 at 11:10pm by DrBob222

Chemistry
A 205 mg sample of a diprotic acid is dissolved in enough water to make 250.0 ml of solution. The pH of this solution is 2.15. A saturated solution of calcium hydroxide (Ksp= 1.3 x 10-6) is prepared by adding excess calcium hydroxide to water and then removing the undissolved ...
Monday, February 25, 2013 at 6:27pm by Summer

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.......
Friday, April 1, 2011 at 1:18pm by DrBob222

Chemistry
a). First, you omitted what you want t calculate!!! .......CH3NH2 + HNO3 ==> CH3NH3^+ + NO3^- mmoles CH3NH2 = 119.2 mL x 0.105 = ?? mmoles HNO3 = 49.1 x 0.255 = ?? Take a look at the numbers; I think the rounded numbers (to 3 s.f.) are equal and the pH (H^+) at the ...
Sunday, February 6, 2011 at 7:23pm by DrBob222

please help analytical chemistry
50 ml of a .1103M solution of formic acid is titrated with a .2511M NaOH. a) What is the pH before and NaOH is added? b) How many mL of NaOH solution must be added to reach the equivalence pt? c) what is the pH at the equivalence pt of the titration? d) If the titration is ...
Monday, April 8, 2013 at 12:24pm by Lana

chemistry
I don't know that you do; however, the problem may be confusing you with the use of pKa (or I suppose it could be confusing me). pK = 7.21 is pK2 for H3PO4 and pK = 12.67 is pK3 for H3PO4. I think you want to use pK2 for H3PO4 and I don't think you need the other one. ...
Sunday, November 14, 2010 at 12:36am by DrBob222

chemistry
50 ml of a .1103M solution of formic acid is titrated with a .2511M NaOH. a) What is the pH before and NaOH is added? b) How many mL of NaOH solution must be added to reach the equivalence pt? c) what is the pH at the equivalence pt of the titration? d) If the titration is ...
Sunday, April 7, 2013 at 6:27pm by Jessica

Chemistry
1.Describe the apparent relationship between H30+ and OH- concentrations when an endpoint is reached in an acid-base titration 2. The indicated end-point of an acid-base titration seldom occurs at a pH of 7. What determines the pH of the end point? 1. They are equal. 2. The ...
Wednesday, May 30, 2007 at 9:50pm by MR

Chemistry!!
Calculate the pH at the equivalence point for the titration of 0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of methylamine is 5.0Ã— 10^â€“4.
Friday, May 11, 2012 at 8:32pm by Sarah

Chemistry
Can you explain what you don't understand here? The first thing you do is calculate where the equivalence point is. A is just calculation of pH of a weak acid. Then you have all of the calculations leading up to the eq. pt, all of the calculations after the eq. pt. and ...
Sunday, February 10, 2013 at 5:35pm by DrBob222

Chemistry
I really don't know how to approach this problem and I really need help. Using the average molarity of your initial acetic acid solutions, the initial volumes, and the volume of NaOH added to reach the equivalence point, calculate the [C2H3O2-] concentration at the ...
Thursday, May 6, 2010 at 2:22pm by Megan

Analytical Chemistry
HBr + NaOH ==> NaBr + H2O 1. First you need to calculate the volume of HBr required to reach the equivalence mols NaOH = M x L = 0.1M x 0.1L =- 0.01 mols HBr must be 0.01 to reach the equivalence point. Then M HBr = mols HBr/L HBr. Substitute to obtain 1M = 0.01/L and L...
Saturday, December 1, 2012 at 3:37pm by DrBob222

chemistry
At the half way point to the equivalence (that would be 22/2 = 11 mL), pH at that point will be pKb
Wednesday, April 4, 2012 at 12:13am 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 ...
Monday, October 1, 2012 at 12:54pm by DrBob222

Chemistry
The formula is pH=pKa+log([A-]/[HA]) A- is the base, HA is the acid. So, Before titration HCl+H2O->H3O+ + Cl- pH=-log[H30+] pH=-log(.100) b/c HCl is a strong acid, it equals [H30+] pH=1 At equivalence point, pH=7
Monday, March 18, 2013 at 8:24pm by Kaylin

CHEM 136
I can't do the entire titration for you. 1. mL acid x M acid = mL base x M base. 2. The secret to a,b,c,d, is in knowing where you are on the titration curve. a. Pure HNO2. Before any NzOH has been added. HNO2 ==> H^+ + NO2^- Ka = (H^+)(NO2^-)/(HNO2) Substitute into...
Tuesday, November 23, 2010 at 4:49pm by DrBob222

chemsitry
1)100ml sample of solution that is 0.2M in both Naf and Hf has 4.0 ml of 1.0M hcl added to it. calculate the ph change of this soultion .ka =7.2*10^-4 2)40 ml of 0.32M benzoic acid if titrated with 60 ml of 0.2 M naoh. clacilate the ph of the resulting solution at the ...
Sunday, March 4, 2012 at 7:24pm by Anonymous

chemistry
1)100ml sample of solution that is 0.2m in both Naf and Hf has 4.0 ml of 1.0m hcl added to it. calculate the ph change of this soultion .ka =7.2*10^-4 2)40 ml of 0.32m benzoic acid if titrated with 60 ml of 0.2 m naoh. clacilate the ph of the resulting solution at the ...
Sunday, March 4, 2012 at 6:28pm by Anonymous

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

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
Titrate the acid as pH vs mL, measure the volume at the equivalence point, take 1/2 that volume, and that pH = pKa.
Saturday, March 30, 2013 at 10:19pm by DrBob222

Chemistry
We can't draw figures on the board. Write the reaction and balance it. At the beginning, the pH is determined by the M of the formic acid. The beginning M is 0.1037 x (25.00 mL/100.0mL) = ? This gives you pH at zero mL base. Next, determine where the equivalence point will...
Monday, March 5, 2012 at 6:10pm 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
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

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!!
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
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
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
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

Chemistry
If 25.0 mL of 0.100 M soln is titrated with 0.10 M base, you know it will take 25.0 mL. So the salt at the equivalence point will be 0.05M (That's 0.025L x 0.1M to begin, you've added 0.025L so the concn is 0.025 x 0.1/0.050 = 0.05M) The pH at the equivalence point is ...
Sunday, February 26, 2012 at 10:44am by DrBob222

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

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 ...
Thursday, October 15, 2009 at 3:35pm by DrBob222

chemistry
Reading the pH at the 1/2 way point to the equivalence oint will give you the Ka if you set pH = -log(H^+) Reading the pH will give you the pKa of the acid directly. Try this. For an acid HA that ionizes HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) at the half way point, (A...
Tuesday, April 8, 2008 at 11:39pm by DrBob222

Science
C6H5NH2 + H2O C6H5NH3+ + OH- 1. Aniline, a weak base, reacts with water according to the rxn above. a. A sample of aniline is dissolved in water to produce 25.0 ml of a .10 M soln. The pH of soln is 8.82. What is Kb for this rxn? b. The soln in part b, is titrated ...
Saturday, April 12, 2008 at 6:27pm by Francis

science
If you are titrating 25.00 mL of a 1.00mg/ml CaCO3 solution, how many mL of 0.010M EDTA solution will be required to reach the equivalence point
Monday, March 4, 2013 at 10:32am by Chris

chemistry
72.33 mL of 0.137 M HCl was added to an Erlenmeyer flask and is titrated with 0.165 M NaOH. Calculate the pH when 5mL of NaOH is added after the equivalence point has been reached.
Wednesday, March 21, 2012 at 7:35pm by blair

Chemistry
A 50.0 mL sample of 0.51 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.40 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 10-5. what am i supposed to do with this?
Sunday, March 8, 2009 at 4:26pm by John

Chemistry
THe pH at equivalence point is always 7
Wednesday, February 4, 2009 at 10:01pm by marsha

chem
what about the pH at equivalence point?
Monday, March 1, 2010 at 11:38pm by simi

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