Search: Calculate the pH after naoh is added

Number of results: 57,428

chemistry
Calculate the pH from the addition of 10 mL of a 0.10 M NaOH solution to 90 mL of 0.10 M HCl. okay so i understand how to do most of it but i get messed up at one part.. so i made my equation: NaOH + HCl -> H2O + NaCl (strong base, strong acid= complete dissociation) so...
Tuesday, November 8, 2011 at 7:12pm by BOB PLEASE HELP!:)

Chemistry/pH- Weak Acid
For a pH = 4, [H+] = antilog(-4.000) = 1.00x10^-4 Ka = [H+][A-] / [HA] Let [H+] = [A-] = x Ka = x^2 / (c - x) , where c = molar concentration of HA before dissociation. if c is much larger than x, we can simplify the expression to: Ka = x^2 / c (1.00x10^-6)c = 1.00x10^-8 c = 1...
Sunday, August 17, 2008 at 6:01am by GK

chemistry
What is the pH of the solution created by combining 12.30 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? mL NaOH pH wHCl pH wHC2H3O2 12.30 What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was...
Monday, May 5, 2008 at 5:31am by Renee

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
There are two problems here. #1. pH when adding 1.90 mL of 0.1 M NaOH to 8.00 mL of 0.1 M HCl. moles NaOH = M x L = 0.0019 x 0.1 = 0.00019 moles moles HCl = 8.00 mL x 0.1 M = 0.008 x 0.1 = 0.0008. NaOH + HCl ==> NaCl + H2O Place the moles below the reactants so you can ...
Sunday, July 11, 2010 at 8:46pm by DrBob222

chemistry help plzzz
HCl + NaOH ==> NaCl + H2O moles HCl = M x L = ?? moles NaOH = M x L = ?? determine which is in excess and calculate H^+ (if HCl is in excess) or OH^- (if NaOH is in excess), and pH from there.
Friday, December 24, 2010 at 5:19pm by DrBob222

AP CHEM
Hello, I'm working these homework questions out; however it is a series of questions and without one step correctly completed, it's nearly impossible to get the others. I posted my work and commented where I need assistance. Thank you so much in advance!!! Potassium ...
Thursday, April 17, 2008 at 8:31pm by Sigurd

Chemistry
Water added to the NaOH dilutes the base and changes the molarity. There are two ways to look at the water added during titration to the Erlenmeyer glask. 1. Water added to the flask dilutes the sample BUT it dilutes the NaOH being added by the same amount. 2. What is the NaOH...
Monday, October 31, 2011 at 10:08am by DrBob222

chem
What is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? I understand how to find the pH for the NaOH + HCl solution. However, I'm not sure how to find it for HC2H3O2. I ...
Monday, May 2, 2011 at 8:10pm by Anonymous

Honors Chemistry
mols NaOH = M x L = ? (NaOH) = (OH^-) = moles/L soln pOH = -log(OH^-) For pH, pH + pOH = pKw = 14. You know pOH and pKw, solve for pH.
Monday, May 21, 2012 at 7:40pm by DrBob222

Analytical Chemistry
Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va = 0,6 and 12 mL
Saturday, December 1, 2012 at 3:37pm by canas

chem101
50.0ml of 0.10M HCL{aq} was added to 40.0ml of 0.10M NaOH{aq} and the mixture was stirred,then tested with a pH meter.What reading should be obtained for pH at 25.0degrees celcias?
Monday, April 5, 2010 at 5:52am by justin

Chemistry
(ii) between 0 and 100.0 mL NaOH added H2A H2O HA - Na+ A2- OH - (iii) 100.0 mL NaOH added H2A H2O HA - Na+ A2- OH - (iv) between 100.0 mL and 200.0 mL NaOH added H2A H2O HA - Na+ A2- OH - (v) 200.0 mL NaOH added H2A H2O HA - Na+ A2- OH - (vi) after 200.0 mL NaOH added H2A H2O...
Friday, March 8, 2013 at 12:13am by Franklin

Chemistry
I took 25 mL of an unknown weak acid and added it to 10 mL of NaOH solution. I measured the pH and got 2.88 with concentration of NaOH @ .0098 and weak acid at 0.0102. What is pKa for the acid?
Saturday, May 5, 2012 at 9:54pm by tc

college chem
A 0.1276 g sample of a monoprotic acid (molar mass = 1.10 x 10^2) was dissolved in 25.0 ml of water and titrated with 0.0633 M NaOH. After 10.0 ml of base has been added, the pH=5.47. What is the Ka for the acid?
Monday, March 28, 2011 at 5:23pm by Angelina

Chemistry
1.0L of aqueous solution in which [H2CO3]=[HCO3^-]=0.10M and has [H^+]=4.2E-7. What is the concentration of [H^+] ofter 0.005 mole of NaOH has been added? H2CO3 ==> H^+ + HCO3^- k1 = (H^+)(HCO3^-)/(H2CO3) I don't know if you are supposed to calculate or to look up ...
Tuesday, August 7, 2007 at 7:26am by Vic

chem
11.80 mL x 0.1M NaOH = 1.180 mmoles. 8.00 mL x 0.1M HCl = 0.800 mmoles 8.00 mL x 0.1M HAc = 0.800 mmoles. ............HCl + NaOH ==> NaCl + H2O initial..0.800...1.180.....0.......0 change..-0.800..-0.800.....0.800.0.800 equi.......0.....0.380.....0.800..0.800 So you ...
Friday, October 28, 2011 at 11:04pm by DrBob222

chemistry
how many grams of NaOH must be added to 0.250 litres of water to form a solution of PH=10.0?
Tuesday, September 7, 2010 at 12:00pm by joey

Chemistry/pH- Weak Acid
Hi again! I have a new question, Can you help me? Consider 50.0 mL of a solution of a weak acid HA (Ka = 1.00.E-6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000? My Calculations: To calculate the concentration of x, I take the pH value -&...
Sunday, August 17, 2008 at 6:07am by Mary

chemistry lab
Could someone help me with this? i am tired of it. here is my data from lab: Buffer1: HPO4- weight 3.412g; original pH Buffer 2: HEPES wt: 1.090g original pH 10.08 buffer1: pH of 0.1M: 7.5; pH of 0.01: 7.72; pH of 0.001M: 7.87 buffer2: PH of 0.1M: 7.5 ph of 0.01 M: 7.42 pH of ...
Friday, February 1, 2008 at 10:54pm by amanda

chemistry
A 20.0 mL sample of 0.687 M HBr is titrated with a 0.216 M NaOH solution. The pH after the addition of 10.0 mL of NaOH is __________. A. 0.413 B. 0.386 C. 0.163 D. 0.579 E. 0.237
Tuesday, July 24, 2012 at 10:44pm by katie

Chemistry
An antacid tablet weighing 1.462 grams was dissolbed in 25mL of 0.8M HCl and diluted with water. The excess HCl was titrated with 3.5 mL of 1.019M NaOH solution. 1. Calculate the number of millimoles added to the tablet 2. Calculate the number of millimoles of NaOH required ...
Monday, October 4, 2010 at 4:12pm by Jennifer

Chemistry
a 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of HCL? what is the ph after the addition of. 0050 mol of NaOH?
Monday, March 7, 2011 at 5:05pm by Anonymous

Chemistry
When 5.00 g of NaOH(s) are added to 100 g of water using a calorimeter, the temperature rises from 25.0 to 37.5 °C. Calculate the molar heat of solution. NaOH(s)--->Na(aq)+OH(aq) Assume that the specific heat capacity of water is 4.18 J/gK and that of the NaOH(aq) ...
Wednesday, November 14, 2012 at 9:08pm by FIrst_lady

science
NaOH + HCl ==> NaCl + HOH strong base + strong acid. Determine mols NaOH = M x L = ?? Determine mols HCl = M x L = ?? Determine which reactant is in excess, then calculate molarity NaOH or HCl and pH from that. Post your work if you get stuck.
Monday, May 5, 2008 at 3:23am by DrBob22

chemistry
HC2H3O2 ==> H^+ + C2H3O2^- Ka = (H^+)(C2H3O2^-)/(HC2H3O2) Calculate (H^+) from pH = -log(H^+). Calculate (C2H3O2^-) from g of the trihydrate. That is moles = grams/molar mass and mole/L (L is 0.15) = M Calculate (HC2H3O2) from L x M = ? Plug in and calculate Ka. For ...
Thursday, July 16, 2009 at 7:35pm by DrBob222

Chemistry
You are preparing 500 mL of a 0.300 M acetate buffer at pH 4.60 using only sodium acetate, 3.00 M HCl, 3.00 M NaOH, and water. Calculate the quantities needed for each of the following steps in the buffer preparation. 1. Add sodium acetate to ~400 mL of water in a large beaker...
Tuesday, February 28, 2012 at 4:28pm by Unknown

chemistry
initial mols HBr = M x L = about 0.014 mols NaOH added = about 0.002 but you need to do them more accurately. ........NaOH + HBr ==> NaBr + H2O I.........0....0.014....0.......0 added....0.002..................... C.......-0.002..-0.002..0.002..0.002 E.........0.....0....
Tuesday, July 24, 2012 at 10:40pm by DrBob222

chemistry
I couldn't guess. How many millimoles acetic acid (HAc) do you have? That's 100 mL x 0.12M = 12. How much NaOH must be added? That's 0.1 M * x mL = 0.1x .........HAc + OH^- ==> Ac^- + H2O initial..12.....0........0.......0 add............0.1x............... ...
Saturday, June 2, 2012 at 7:29pm by DrBob222

Chem Webwork UCI
Complete the table below: What is the pH of the solution created by combining 0.70 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)? mL NaOH: 0.70 pH w/ HCl: 1.08 pH wHC2H3O2: 3.23 Complete the table below: What are the pH ...
Monday, April 25, 2011 at 4:50pm by Michelle

Chemistry
Calculate the pH after 0.010 mol gaseous HCl is added to 250.0 mL of each of the following buffered solutions: a. .050 M NH3/.15 M NH4Cl b. .5 M NH3/ 1.5 M NH4Cl
Wednesday, April 3, 2013 at 9:04pm by Joyy

chem
I didn't see that the pH at the equivalence point was in the question. I just looked again and it isn't. But the pH at the equivalence point is determined by the hydrolysis of the salt. Bu^- + HOH ==> HBu + OH^- Kb = (Kw/Ka) = (OH^-)(HBu)/(Bu^-) I think Kb is ...
Monday, March 1, 2010 at 11:38pm by DrBob222

chemistry
NaOH + HCl ==> H2O + NaCl mols HCl = M x L = ? mols NaOH = M x L = ? Subtract, the excess will b either acid or base. If acid, then pH = -log(HCl) = ? If base, then pOH = -log(NaOH), then pH + pOH = pKw = 14. Solve for pH.
Thursday, May 10, 2012 at 12:27am by DrBob222

chemistry
a buffer composed of 0.50mol acetic acid and 0.50mol sodium acetate is diluted to a volume of 1.0L. The pH of the buffer is 4.74. How many moles of NaOH must be added to the buffer solution to increase its pH to 5.74?
Wednesday, October 7, 2009 at 9:20pm by millie

Chemistry
A volume of 25.0 ml of 0.100 M CH3C02H is titrated with 0.100 M NaOH. What is pH after the addition of 12.5 ml of NaOH? (Ka for CH3CO2H = 1.8x10-5)
Monday, October 18, 2010 at 11:44pm by Helen

chemistry
Calculate the pH in a 0.00500 solution of NaOH.
Monday, February 21, 2011 at 11:15pm by JC

Chemestry
A 1L of buffer was prepared from 0.15 mole of Na2HPO4 and 0.10 moles KH2PO4. a) What is your pH and what is the pH change that adding of 80.0 mL of 0.100 mol L-1 NaOH cause? b) What happen to the pH after adding 10.0 mL of 1.0 mol L HNO3-1.
Thursday, April 28, 2011 at 10:43am by Margo

chemistry
For the titration of 50.00 mL of 0.1000 M ammonia with 0.1000 M HCl, calculate the pH (a) before the addition of any HCl solution, (b) after 20.00 mL of the acid has been added, (c) after half of the NH3 has been neutralized, and (d) at the equivalence point.
Wednesday, April 11, 2012 at 6:03am by chamy

Chemistry, acids and bases
(a) Find pH and pOH of 1.0 M solution of sodium benzoate, NaC6H5COO. (Ka of benzoic acid is 6.2x10^-5) (b) Calculate pH after 0.205 moles per liter of HCl was added (assume volume does not change (c) Calculate pH after 1.0 moles per liter of HCl was added (assume volume does ...
Tuesday, May 6, 2008 at 8:45pm by Marissa

Chemistry
millimoles HCl = mL x M = 100 x 0.000018M = 0.0018. millimoles NaOH added = 1 mL x 0.1M = 0.1. ..........HCl + NaOH ==> NaCl + H2O initial..0.0018..0.1.......0......0 change..-0.0018.-0.0018.+0.0018.0.0018 equil....0......0.0982...0.0018.0.0018 You can see that all of ...
Sunday, December 11, 2011 at 8:37am by DrBob222

AP Chem
Calculate the pH after 0.010 mol gaseous HCl is added to 250.0 mL of each of the following buffered solutions: a. .050 M NH3/.15 M NH4Cl b. .5 M NH3/ 1.5 M NH4Cl
Wednesday, April 3, 2013 at 2:48pm by Kristen

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

chemistry repost
Alex--Is there any/some indication of the end point to which the NaOH/Na2CO3 was titrated? My point is that Na2CO3 to the phenolphthalein end point (about pH 9) titrates to NaHCO3 whereas Na2CO3 titrated to the methyl red or methyl orange end point (about pH 4 or 5) titrates ...
Monday, April 21, 2008 at 9:38pm by DrBob222

Chemistry
By adding NaOH you have formed a buffered solution. Take 1 L soln, so you have 1000 x 0.559 M = 559 millimols PhCOOH. 1000 x 0.0225M = 22.5 mmols NaOH .......PhCOOH + NaOH ==> PhCOONa + H2O initial..559................0........0 added...........22.5................... ...
Tuesday, May 8, 2012 at 4:51pm by DrBob222

science(chem)
I know that the solution was acidic after rxn since I added sodium bicarbonate after testing for the pH
Monday, April 7, 2008 at 11:00pm by ~christina~

Chemistry
A .682 g sample of an unknown weak monoprotic acid, HA, was dissolved in sufficient water to make 50mL of solution and was titrated with a .135 M NaOH solution. After the addition of 10.6mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the ...
Thursday, May 10, 2007 at 2:57pm by Sara

chemistry
I assume you are using pKa = 4.74 so that pH = 4.74 + log 0.5/0.5 = = 4.74 + log 1 = 4.74 + 0 = 4.74. If we call acetic acid HAc and the acetate ion (the base) Ac, then To make the pH 5.74 we calculate the B/A as follows: 5.74 = 4.74 + log B/A You can go through the ...
Wednesday, October 7, 2009 at 9:20pm by DrBob222

Chemistry
A generic equation can be written as follows: RCOOH + NaOH ==> H2O + RCOONa mols NaOH = L x M mols RCOOH = mols NaOH (since the acid is monoprotic). mols RCOOH = grams/molar mass. You have mols and grams. Solve for molar mass. The pH of 4.92 at 17.54 mL NaOH is ...
Sunday, June 8, 2008 at 9:47pm by DrBob222

chemistry
how do i calculate the NaOH molarity from a titration of KHP and NaOH after preparing KHP to standardise NaOH. KHP mass was 2.0378g dissoveld in distilled water to fill up a 100ml volumetric flask and only 10ml of this solution was used to in titration
Friday, March 22, 2013 at 7:00am by k

Analy Chem
Consider the titration of 100.0 mL of 0.110 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va:Va = 0,2,5,8,10,10.1,11,11.1, and 12mL. Compute Ve, first.
Wednesday, April 11, 2012 at 6:54pm by curtis

AP Chemistry
i found the pH to be 9.75, i don't know if i did it right, Assuming i did it right, there is another question stating what is the pH of the buffer after 90.0mL of 2.00M HCL are added?...
Sunday, April 11, 2010 at 3:51pm by Yamani

Chemistry HELP!
for a). notice that L x M = moles and moles x molar mass = grams; therefore, L x M x molar mass = grams and molar mass = g/(L*M) so M too low means molar mass is too high. b). Did you use the pH meter to record the pH after drop wise addition of NaOH, then plot the data to ...
Wednesday, April 13, 2011 at 4:31pm by DrBob222

AP CHEM
Potassium hydrogen phthalate is a weak solid acide w/ the formula KHc8H4O4 (often abbrv. KHP) and a formula weight of 204.23 g/ mole. It is often used to standardize a solution of a strong acid with an unknown concentration(Actually KHP is used to standardize a base and that ...
Thursday, April 17, 2008 at 8:31pm by DrBob222

chemistry
The only way I know to do this is to calculate the buffer capacity of each, then compare them. The buffer capacity is the number of moles of a strong acid or a strong base that causes 1.00 L of the buffer to undergo a 1.00 unit change in pH. So you calculate the pH of the ...
Sunday, December 12, 2010 at 5:28pm by DrBob222

Chemistry
Calculate the pH of 30.0 mL of 0.025 M NaOH
Tuesday, November 27, 2012 at 12:35am by Jen

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
How many milliliters of 1.0 M NaOH must be added to 200. mL of 0.10 M NaH2PO4 to make a buffer solution with a pH of 7.50? (5 points)
Thursday, April 4, 2013 at 7:27pm by bill

General Chemistry
How many milliliters of 0.5M NaOH must be added to 200 mL of 0.1M NaH2PO4 to make a buffer solution with a pH of 6.9?
Thursday, January 31, 2013 at 11:05am by Joanna

AP Chemistry
There is an unknown amount of unlabelled monoprotic acid in an unknown amount of water titrated with a sample with a solution of NaOH of unknown molarity. After adding 10.0 mL of NaOH, the pH=5.0. The equivalence point is 32.22 mL of NaOH. What is the Ka? I have been looking ...
Tuesday, March 8, 2011 at 9:58pm by Ben

Chemistry Lab
HA + NaOH ==> NaA + H2O mols NaOH = M x L = ? mols HA = mols NaOH (look at the equation. 1 mol HA = 1 mol NaOH) M HA = mols HA/L HA. .........HA ==> H^+ + A^- I........?M......0......0 C.........-x.....x.......x E........?M -x...x.......x Ka = (H^+)(A^-)/(HA) ...
Tuesday, October 30, 2012 at 1:07pm by DrBob222

chemistry
I will do the first one and you can do the others. pH = pKa + log[(base)/(acid)] I found Ka of 7.2E-4 for HF but you need to use the number in your text. pKa from this value is 3.14. pH = 3.14 + log(NaF/HF) pH = 3.14 + log(0.4545/0.4105) = 3.18 [Note: I have divided the ...
Thursday, November 25, 2010 at 5:40pm by DrBob222

Science-Chemistry
Methyl amine is a weak base with a pKb=3.35. Consider the titration of 30.0mL of .030M of Methyl amine with 0.025M HCl. a) Write the appropriate equation for the reaction. b) Calculate K for the reaction in part (a). c) Calculate pH of the initial methyl amine solution (0mL ...
Friday, April 19, 2013 at 11:44am by Joe

Chemistry
A buffer is made by mixing 52.1 mL of 0.122 M acetic acid with 46.1 mL of 0.182 M sodium acetate. Calculate the pH of this solution at 25 degrees C after the addition of 5.82 mL of 0.125 M NaOH
Saturday, May 1, 2010 at 12:48pm by Chemistry Problem

Chemistry
What molarity of NaOH would produce a pH of 10? And how do you calculate it?
Wednesday, April 16, 2008 at 9:26am by Ashley

Chemistry
A buffer is prepared by mixing 45.0 mL of 0.15 M NaF to 35.0 mL of 0.10 M HF (Ka = 6.8x10^-4) Calculate the pH of this buffer after 5.0 mL of 0.05 M HCl is added.
Sunday, April 17, 2011 at 4:20pm by Sam

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

chemistry
From the pH. pH = -log(H^+). You know pH, calculate (H^+), then use your CV formula. Note, however, that the questions asks what volume of water is to be ADDED (not what is the final volume). Do your CV formula to find the total volume, then subtract from 1.00 L to find the ...
Tuesday, August 4, 2009 at 9:43pm by DrBob222

Chemistry
Use the Henderson-Hasselbalch equation. 5.10 = 4.74 + log(base/acid) Calculate ratio base/acid. You know you have added 2 mmoles NaOH which will allow you to calculate acid at the pH = 5.10 point. Determine the amount mL required to neutralize that many mmoles acid, add to the...
Tuesday, November 9, 2010 at 10:34am by DrBob222

Chemistry
I find it helpful to write each step of the procedure and see how the question affects each. 1. mols KHP = grams/molar mass 2. mols NaOH = mols KHP 3. M NaOH = mols NaOH/L NaOH So you did steps 1 and 2 right but on 3 added too much L NaOH. That's in the denominator, too ...
Wednesday, December 5, 2012 at 3:17pm by DrBob222

Chem titration ph
What arethe pH, pOH, and concentrations of CO3 2- and HCO3 - after 1.000 mL of 0.1500 M HCl is added to 100.0 mL of 0.0100-M sodium carbonate solution? ignore the added1% volume. CO3^-2 + H^+ ==> HCO3^- Calculate mols carbonate to start. Calculate mols HCl to start. ...
Thursday, October 5, 2006 at 3:00pm by Andy

Chemistry
No, I'm sorry I didn't see it. Those get buried after a page or two. You added 75.0 mL HCl initially. It reacted with the tablet and used up all of its neutralizing power. So mmoles HCl added initially = mL x M which we've done. But you had some HCl left over ...
Monday, October 18, 2010 at 10:04pm by DrBob222

chemistry
Calculate the pH range required to separate 99.99% of Cr3+ from Zn2+ by precipitation of Cr(OH)3(s) in a solution that is initially 0.034 M in both Cr3+ and Zn2+. Use the Ksp values from Table 16.2 of Chang and calculate the pH to the nearest 0.01 pH unit. Lowest pH= Highest pH=
Sunday, March 25, 2012 at 5:17pm by Tanner

AP Chemistry
In the titration of 50.0 mL of 1.0 M CH3NH2 (kb=4.4 x 10^-4), with 0.50 M HCl, calculate the pH a) after 50.0 mL of 0.50 M has been added b) at the stoichiometric point.
Sunday, March 29, 2009 at 11:41pm by some kid

Chemistry
Convert 0.40 g NaOH to moles. Convert 50.0 mL x 0.1 M HCOOH to moles. See which is in excess. If NaOH, then calculate pOH and pH. If HCOOH, it will be a buffer and you should use the Henderson-Hasselbalch equation.
Wednesday, May 5, 2010 at 12:28pm by DrBob222

chemistry
ahhh im confused!! Calculate the pH of a solution formed by mixing 372 mL of a solution containing 4.4 x 10-6 M NaOH with 286 mL of 6.0 x 10-2 M NaOH. Report your answer to 2 decimal places.
Friday, November 16, 2007 at 6:14pm by mark

chemistry
Draw the tritration curve for the neutralization of a 100ml sample of 1M solution of HCl with a .6M solution of Naoh point 1: 2 ml NaOH edded point 2: 40 ml NoOH ADDED POINT 3: EQUIVALENCE POINT 4: 80 ML NaOH added
Sunday, April 3, 2011 at 1:55pm by sam

Chemistry
.012 mol of solid NaOH is added to a 1 L solution of HCl 0.01 M. What is the pH of the solution?
Monday, May 13, 2013 at 7:17pm by scilover

chemistry
The problem statement, all variables and given/known data When 5.00 g of NaOH(s) are added to 100 g of water using a calorimeter (with Cp = 493.24 J/K), the temperature rises from 25.0 to 37.5 °C. Calculate the molar heat of solution. NaOH(s)--->Na(aq)+OH(aq) Assume...
Wednesday, November 14, 2012 at 4:16pm by fufi

Chemistry
A .288 g sample of an unknown monoprotic organic acid is dissolved in water and titrated with a .115 M sodium hydroxide solution. After the addition of 17.54 mL of base, a pH of 4.92 is recorded. The equivalence point is reached when a total of 33.83 mL of NaOH is added. What ...
Sunday, June 8, 2008 at 9:47pm by johnny

Science (Chemistry)
I'm not sure how to do a couple of problems on my chemistry worksheet. Can you please help me? You dissolve 0.00902 g of naOH in enough water to make 1000 mL. What is the Ph of this soultion. What is the molarity, pH, and pOH of 5.00 of naOH in 750.0 mL of soultion. I know...
Sunday, November 6, 2011 at 10:18am by Dylan

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

Chemistry
1. Calculate the pH of a buffer solution that contains 0.32 M benzoic acid (C6H5CO2H) and 0.17 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid] Round your answer to two places past the decimal. 2. A solution is prepared by mixing 470 mL of 0.18 M Tris&...
Monday, April 8, 2013 at 1:31am by Erin

chemisty
There are two ways, one is an estimate and the other is exact. For the estimate, look at the pH values. When the pH changes from low numbers (you said you were titrating HCl by adding 1 mL portions of NaOH) to high numbers (pH 2 or 3 to pH 9 or 10), the equivalence point is ...
Monday, February 6, 2012 at 3:31pm by DrBob222

Chemistry
Part A: Unknown Acid use 1gram and mix with 120 mL of distilled water Determine the concentration of the acid by titrating with 0.0998 M of NaOH Concentration of NaOH= 0.0998M Volume of NaOH= 3.5 mL # moles of NaOH= Initial Concentration of Weak Acid- Show Calculation of Ka ...
Wednesday, March 24, 2010 at 3:40pm by Saira

Chemistry
NaOH ==> Na^+ + OH^- NaOH is 0.04M from the problem. Na^+ is 0.04M and OH^- is 0.04M You took the -log(OH^-), which is 1.40 and you called that the H^+ but it isn't. It is the (OH^-) so you calculated the pOH and not the pH. But you can convert your pOH to pH from ...
Thursday, December 23, 2010 at 9:54pm by DrBob222

Whoops
[pH][pOH] = 1 x 10^-14 pH = 1 x 10^-14 / 9.69 x 10^-12 No, pH = 1 x 10^-14 / 9.69 x 10^-2 = 1.03 x 10^-13 about Ph of 13, which is about 4% NaOH You know the Ph has to be much higher than 7, nearly 14, because NaOH is a BASE !
Monday, July 27, 2009 at 1:32am by Damon

Chemistry using concentration
Then you didn't read it very carefully. I'll draw you a picture. mols NaOH = grams/molar mass You have grams NaOH of 1.852. You can get molar mass NaOH. Calculate mols NaOH. Calculate mols NaOH needed.The above does that calculation for you but you must plug in the ...
Monday, May 12, 2008 at 12:37am by DrBob222

Chemistry
For a. You have 0.00 mL of NaOH; therefore the pH will be determined solely by HCl. pH = -log(0.250) pH = 0.602
Saturday, April 27, 2013 at 3:02am by DrBob222

chemistry
A volume of 25.0 mL of 0.100 M CH3CO2H is titrated with 0.100 M NaOH. What is the pH after the addition of 12.5 mL of NaOH? (Ka for CH3CO2H = 1.8 x 10-5)
Tuesday, May 31, 2011 at 7:37am by Al

college
A volume of 25.0 ml of 0.100 M CH3CO2H is tiltrated with 0.100 M NaOH. What is pH after the addition of 12.5 mL of NaOH? (Ka for CH3CO2H = (1.8 x 10-5)
Wednesday, October 20, 2010 at 12:22am by Helen

chemistry
Use the Henderson-Hasselbalch equation. Construct an ICE chart. Add 20 mL of 0.75M (I guess that's M but you omitted the symbol) NaOH = 15 millimoles NaOH = 0.015 mols. (Note: I simply assume the NaOH was not present initially and it was added later. That doesn't ...
Monday, March 19, 2012 at 10:25am by DrBob222

chemistry
Blank corrections are useful in correcting titrations in which some unwanted side reactions occur. IN this case, measure the amount of ethanol used in the experiment, add indicator, and titrate with the NaOH until the indicator changes color. The mL NaOH added is the blank ...
Saturday, January 8, 2011 at 2:45pm by DrBob222

Chemistry
Consider the titration of 41.2ml of 0.230 M HF with 0.220 M NaOH . Calculate the ph after the addition of 11.9ml of sodium hydroxide.(HF: Ka = 3.50 x 10-4) Please, I need step by step on on how to solve this type of problem. Thank you very much!
Tuesday, April 23, 2013 at 1:22am by Chloe

College Chemistry
1) A 25mL sample of the .265M HCI solution from the previous question is titrated with a solution of NaOH. 28.25mL of the NaOH solution is required to titrate the HCl. Calculate the molarity of the NaOH solution. 2) A 1.12g sample of an unknown monoprotic acid is titrated with...
Monday, October 25, 2010 at 2:29pm by Jessica

chemistry
what is the pH of the solution formed when 25 mL of 0.173 M NaOH is added to 35 mL of 0.342 M HCl?
Wednesday, December 5, 2012 at 11:30am by sarah

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