Search: how many ml of 0.25 M NaOH needed to titrate 75 ml 0.1 M HCl

Number of results: 36,314

Chemistry
25 mL 0.1 M acetic acid is titrated with 0.1 M NaOH. Calculate the pH after the addition of 0.0 mL, 1.0 ml, 2.5 mL, 4.0 mL, 7.0 mL, 10.5 mL, 15.0 mL, 17.0 mL, 20.0 mL, 22.0 mL, 23.0 mL, 23.5 mL, 24.0 mL, 24.25 mL, 24.5 mL, 24.75 mL, 25.0 mL, 25.25 mL, 25.5 mL, 25.75 mL, 26.0 ...
Monday, April 22, 2013 at 4:34am by smartkid <33

Chemistry
Look at titrating the first H, then double it. H3PO4 + NaOH ==> NaH2PO4 + H2O mols H3PO4 = 25 mmols. So you will need 25 mmols NaOH to react with the first H of H3PO4. M NaOH = mols NaOH/L NaOH 2.0M = 25 mmols/mL mL = 25/2.0 = 12.5 mL. So it will take 25 to neutralize ...
Saturday, March 23, 2013 at 5:32pm by DrBob222

AP CHEMISTRY
mL NaOH x M NaOH = mL acid x M acid. mL NaOH x 0.125M = 25.0 x 0.175 Solve for mL NaOH. The 20% ionization is extraneous information.
Friday, March 29, 2013 at 1:20am by DrBob222

Chem
For the first one, check your work. I think you have an error of about 1000 times. I think 500*0.200/0.150 = 667 mL. However, that is the volume you will have, not how much water must be added. Water that must be added is 667-500 = ?? For the second one, M*mL = M*mL. 125 mL * ...
Friday, May 1, 2009 at 12:20am by DrBob222

Chemistry
Which of the following solution should be mixed with 50.0 mL of 0.050 M HF solution to make an effective buffer? A) 50.0 mL of 0.10 M NaOH B) 25.0 mL of 0.10 M NaOH C) 50.0 mL of 0.050 M NaOH D) 25.0 mL of 0.050 M NaOH
Sunday, April 1, 2012 at 9:20am by Anon

Chemistry
Calculate the pH of a 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCL. (0.1 M HCl and NaOH) In total, you need to show 10 calculations.
Monday, March 14, 2011 at 9:32pm by Mariah

college chem
Calculate the pH of a 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCL. (0.1 M HCl and NaOH) In total, you need to show 10 calculations.
Monday, March 14, 2011 at 9:50pm by Mariah

Chemistry
Which one of the following pairs of 0.100 mol L-1 solutions, when mixed, will produce a buffer solution? A. 50. mL of aqueous CH3COOH and 25. mL of aqueous CH3COONa B. 50. mL of aqueous CH3COOH and 25. mL of aqueous HCl C. 50. mL of aqueous NaOH and 25. mL of aqueous HCl D. 50...
Sunday, May 2, 2010 at 11:27pm by Macky

chem 2
which one of the following pairs of 0.100 mol L-1 solutions, when mixed will produce a buffer solution? a 50. mL of aquesous CH3COOH and 25.mL of aquesous HCI b 50. mL of quesous CH3COOH and 100. mL of aqueous NaOH c 50. mL of aquesous NaOH and 25. mL of aqueous HCI d 50. mL ...
Saturday, March 3, 2012 at 1:42pm by leaha

chemistry
Can you help me with the following: Calculate hte pH that results when 25 mL of .1M HCL is titrated with .1M NaOH solution run in from a buret at each of the stages of volume of .1 M NaOH add in 5.0 mL incrememnts beginning with 0 mL of NaOH
Friday, April 30, 2010 at 12:08pm by Cathy

Chemistry
So far so good; however, you didn't include the volume of NaOH required in the titration. M(NaOH) x mL(NaOH) = M(vinegar) x mL(vinegar). 0.1250 x mL(NaOH) = M(vinegar) x 4.00 mL Plug in the mL NaOH, whatever it is, and you will have just one unknown which is the molarity ...
Wednesday, July 23, 2008 at 10:44pm by DrBob222

Chemistry - Molarity
I think something is missing but I don't know what. KHP + NaOH ==> H2O + NaKP H2P + 2NaOH ==> 2H2O + Na2P -------------------------- Suppose the sample is 100% KHP, then 25.0 mL x 0.127M = 3.175 mmols. That will take 3.175 mmols NaOH M NaOH = mmols/mL; ...
Monday, December 10, 2012 at 10:56pm by DrBob222

CHEM 102
Can someone help me with this question please? What is the pH at each of the points in the titration of 25.00 mL of 0.1000 MHCl by 0.1000 M NaOH:i) After adding 25.00 mL NaOH ii) After adding 26.00 mL NaOH thankyou
Thursday, August 18, 2011 at 11:40pm by paige

chemistry
Hello, im having major issues with this problem...A 100.0-mL solution of 0.017 M CH3COOH (Ka = 1.8 10-5) is titrated with 0.025 M NaOH. Now, I've gotten some the answers all ready (right answers have *)....but three of them are wrong, can someone tell me what to do? 1.) ...
Wednesday, October 12, 2011 at 9:44pm by A person

chem-please help!!!!!
when a 25.0 mL sample of an unknown acid was titrated with a 0.100 M NaOH solution. Determine Ka for the unknown acid. Volume NaOH = 25 mL; pH=8.25
Monday, October 18, 2010 at 1:12pm by Rosemary

Chemistry
I assume this is 25% w/v. 25% w/v citric acid means 25 g citric acid in 100 mL soln or 250 g citric acid in 1000 mL. Convert 250 g citric acid to moles (moles = grams/molar mass). 2. Write the equation. 3. Using the coefficients in the balanced eqaution, convert moles H^+ in ...
Friday, August 5, 2011 at 11:24am by DrBob222

Chemistry
M NaOH x mL NaOH = M HClO4 x mL HClO4 0.300M x mL NaOH = 0.15M x 30.0 mL You have one unknown; i.e., mL NaOH. Solve for that.
Wednesday, July 18, 2012 at 10:03pm by DrBob222

Chemistry
What are the factors that can cause titration of 25 mL mixture of sodium bicarbonate and sodium carbonate with HCL not to reach the end point especially after addition of two 25 mL aliquots of NAOH and 10ml BACO3? And why is it that 25ml aliquots of NAOH is used rather than 50...
Saturday, March 7, 2009 at 4:20am by chemy

chemistry
Hello, im having major issues with this problem...I need to find a pH for a 100.0-mL solution of 0.017 M CH3COOH (Ka = 1.8 10-5) is titrated with 0.025 M NaOH. Now, I've gotten some the answers all ready (right answers have *)....but three of them are wrong, can someone ...
Wednesday, October 12, 2011 at 11:20pm by A person

chemistry
20 ml of 0.1 M Acetic acid and 25 ml of 0.1 M Sodium Acetate are mixed together. then 5ml of 0.1 M of NaOH is added. Find the Ph before the addition of NaOH and after the addition of NaOH??
Tuesday, March 15, 2011 at 12:54am by james bond

chem
20 ml of 0.1 M Acetic acid and 25 ml of 0.1 M Sodium Acetate are mixed together. then 5ml of 0.1 M of NaOH is added. Find the Ph before the addition of NaOH and after the addition of NaOH??
Tuesday, March 15, 2011 at 12:41am by james bond

Chemistry
c1v1 = c2v2 c = concn v = volume For example. You have 2M NaOH and you want to make 500 mL of 0.25M. How much of the 2M stuff do you take? 2M*x mL = 0.25M*500 mL mL = (0.25*500/2) = 62.5 mL So you take 62.5 mL of 2M NaOH, add to 500 mL volumetric flask and make to the mark. ...
Wednesday, January 16, 2013 at 10:41pm by DrBob222

Chemistry
Titration of 25.00 mL of a monoprotic acid solution with 0.2269 M NaOH is: Initial NaOH = 3.96 mL Final NaOH = 38.84 mL (dif is 34.88mL) Based on data what is the molar concentration of acid? I did V x M = 0.03488L NaOH x 0.2269 = 0.079143 Not sure if I should have switched to...
Friday, December 7, 2012 at 11:44am by Terry

Chemistry
I am assuming that the problem does not mean that mL acid + mL base = 500 mL. With that assumption, we see that a starting quantity for neither formic acid nor NaOH is given; therefore, an infinite number of answers are possible depending upon how much formic acid we take ...
Wednesday, September 26, 2012 at 11:13am by DrBob222

chemistry
How many milliliters of 0.729 M NaOH are needed to prepare 150.0 mL of 0.122 M NaOH? 25.1 mL
Wednesday, February 22, 2012 at 6:34pm by krystal

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

Chem
25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH. my work CH3COOH + H2O <-> H3O^+ + CH3COO^- 25mL x 0.100 mmol/ml = 2.5mmol CH3COOH 27mL x 0.100 mmol/ml = 2.7mmol NaOH ...
Tuesday, June 29, 2010 at 4:02pm by Anonymous

chem
What volume of a 0.2 N NaoH solution do you expect to use in order to titrate 25ml (sample out of 250ml) of a 0.200 N oxalic acid solution? heeeelp! Equation is H2C2O4 + 2NaOH==> 2H2O + Na2C2O4 I assume that is 25.00 mL H2C2O4. milliequivalents(meq) NaOH = meq H2C2O4 25...
Tuesday, September 26, 2006 at 4:00pm by Melba

General Chemistry
millimoles HCl = 8.00 mL x 0.1M=0.8mmols. mmoles NaOH = 2.50 mL x 0.1M = 0.25 mmols. mmoles NaOH = 9.50 x 0.1M = 0.95 mmoles. mmoles HAc = 8.00 mL x 0.1M = 0.8 mmoles. -------------------------------------- ..........HCl + NaOH ==> H2O + NaCl initial...0.8....0........0...
Sunday, November 6, 2011 at 9:28pm by DrBob222

chemistry
First calculate the molarity of the NaOH. It is diluted 25 mL to 1,025 mL. So 2.0 M x (25/1025) = ?? NaOH is a strong electrolyte and is ionized 100%; therefore, OH^- = the molarity. Take it from here with the following: pOH = - log (OH^-) pH + pOH = 14.
Thursday, November 1, 2007 at 10:09pm by DrBob222

AP CHEMISTRY
Well, it turns out that way and to be a little honest about it I was surprised. But here are two ways to do it. Let's assume the volume of the acid is ANYTHING but for the first case we'll take 50 mL. The 0.1M x 50mL = 0.08MNaOH x ?M. Solve for ?M NaOH which is the ...
Thursday, March 1, 2012 at 4:46pm 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
mols HCl = M x L. M you have as 0.2500. L. Final 25.20mL - 2.08 mL = mL HCl. Convert to L and find mols as above. Then mols NaOH = mols HCl. M NaOH = mols NaOH/L NaOH. Solve for M.
Wednesday, November 28, 2012 at 7:32am by DrBob222

chemistry
A volume of 100mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after...
Monday, August 6, 2012 at 8:00pm by Anon

Chemistry
A 25 mL sample of wine is found to have a concentration of acetic acid of 0.23 M. If this was titrated with a 0.16 M NaOH solution how many mL's of NaOH would be required?
Saturday, November 27, 2010 at 5:44pm by Susan

chem
calculate the pH during the titration of 100 mL of .200 M HCl with .400 M NaOH after 0,25, 50, and 75 mL NaOH have been added
Wednesday, March 28, 2012 at 10:03pm by Brynn

chemistry
A student needed to standardize a solution of NaOH which was approximately 0.125 M. The student carefully prepared the titration setup, but after 25 mL of NaOH was added, what was the pH of solution if initially flask contains 25 mL of 0.250 M HCl solution?
Sunday, October 17, 2010 at 6:51pm by Cindy

CHEMISTRY
A student needed to standardize a solution of NaOH which was approximately 0.125 M. The student carefully prepared the titration setup, but after 25 mL of NaOH was added, what was the pH of solution if initially flask contains 25 mL of 0.250 M HCl solution?
Sunday, October 17, 2010 at 4:50pm by Maria

Chemistry
One example each for 0.1M HCl/NaOH added to 25 mL H2O. The diluted HCl solution is 0.1M x (1 mL/26 mL) = (HCl) pH = -log(HCl) The diluted NaOH solution is 0.1M x (1 mL/26 mL) = (NaOH) pOH = -log(NaOH). Then pH + pOH = pKw = 14.00
Monday, March 14, 2011 at 9:32pm by DrBob222

General Chemistry
"How would you prepare 150. mL of 0.1M sodium hydroxide given a stock solution of 3.0M NaOH?" I made calculations and figured the process would be to dilute 5 mL of the 3.0M NaOH to 150. mL. Is this correct? For the ideal 150. mL of 0.1M NaOH, I calculated there to ...
Sunday, April 21, 2013 at 9:18pm by Emily

Chemistry
I can show you how to solve the problem but I have no numbers. However, mL vin x M vin = mL NaOH x M NaOH If you substitute mL vinegar and multiply by M vinegar (molarity = mols/L) and divide by M NaOH you can calculate mL NaOH. Most vinegars are about 5% acetic acid and I ...
Saturday, April 27, 2013 at 7:50pm by DrBob222

Chemistry
It requires 25.1 mL of 0.110 M HCl to titrate 12.0 mL of an NaOH solution of unknown concentration. What is the NaOH concentration?
Monday, February 27, 2012 at 7:07pm by Kyle2

Chemistry
How much volume(mL)of 2.50 M NaOH(aq)must be added to 25.0 mL of water to produce a 1.50 M NaOH solution? Thanks!
Monday, April 23, 2012 at 11:37am by Jack

Chemistry
My calculator is on the blink; I can't do a calculation. H2SO4 + NaOH ==> H2O + NaHSO4 NaHSO4 + NaqOH ==> H2O + Na2SO4 50 mL H2SO4 x 0.5M = 25 millimoles. 50 mL NaOH x 0.50M = 25 mmols. Therefore, the first H must have been replaced and you are left with ...
Monday, April 8, 2013 at 9:45am by DrBob222

Chemistry
Exactly 23.6 ml of a 0.131 N HCl solution was required for complete neutralization of 25.0 ml of an NaOH solution.What was normality of the NaOH?
Saturday, June 25, 2011 at 4:36am by Rajan Khare

Chemistry
A 50.0-mL sample of 0.10 M HNO2 is titrated with 0.10 M NaOH. What is the pH after 25.0 mL of NaOH have been added?
Tuesday, December 7, 2010 at 8:50pm by Eric

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
I there, this is my lab worksheet, I am having trouble filling it out. Please help me with it. Expt #1- Molecular Weight of Unknown Acid Unknown Acid: #3 Mass of Unknown solid acid transferred:1.0g Volume of volumetric flask: 100.00 mL Concentration of NaOH: 0.0989 M Aliqot of...
Monday, October 4, 2010 at 10:17pm by Steve

College Chemistry
I need help with this question!! a sample of pure KHP weighing .8097g dissolved in water and titrated with 40.25 mL of NaOH solution. The same NaOH solution was used to titrate a solution of a weak diprotic acid H2X. A sample of 0.18694g of the weak acid was first dissolved in...
Wednesday, April 25, 2012 at 7:43pm by Emily

chemistry
What volume of WATER MUST BE ADDED to 10.0 mL of 6.0 M NaOH to make a solution that is 0.30 M in NaOH? Assume that the volumes are additive. a. 10 mL b. 190 mL c. 200 mL d. 210 mL e. 500 mL
Monday, March 29, 2010 at 6:30am by lawrence

Chemistry
Hi,may I have your help? I have posted a titration question that I'm trying to solve. The correct answer to the problem is 0.128 M, however,I have arrived at the answer: 0.000128 M. Here's the question: If 38.30 mL of 0.250 M NaOH is used to titrate 25.0 mL of ...
Friday, December 10, 2010 at 5:37pm by Cliff

chemistry
When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution increases to 33.9 °C. Assuming the specific heat of the solution is 4.18 J/(g·°C), that its density is 1.00 g/mL, and that the ...
Thursday, March 14, 2013 at 7:57pm by nobody

CHEMISTRY
Suppose a student adds 25.00 mL of 1.025 M HCl to a 1.50 g antacid tablet. The student boils and then titrates the resulting solution to the endpoint with 0.4969 M NaOH. The titration requires 21.3 mL NaOH to reach the endpoint. How many moles of HCl were neutralized by the ...
Tuesday, April 23, 2013 at 12:40am by Jo

Chemistry
What is the normality of 10 ml of an acid that took 25 ml of 0.25 N NaOH to reach pH 7?
Wednesday, February 17, 2010 at 1:10pm by David

Chemistry
moles NaOH = M x L = 0.613 x 0.03375= ?? So moles acidity in the solution (the vinegar) is the same as moos NaOH since the reaction is a 1 mole to 1 mole ratio. Since moles vinegar is in 25 mL, then M = moles/L and M vinegar = moles from above/0.025 = ?? OR, if you don't ...
Tuesday, June 22, 2010 at 4:44pm by DrBob222

ap chemistry
please explain titration problems. I'm a total noob at this and am trying to answer some prelab questions. Examples: 1. How many mL of a 0.800 M NaOH solution is needed to just neutralize 40 mL of a 0.600 M HCl solution? 2. You wish to determine the molarity of a solution ...
Thursday, October 2, 2008 at 10:19pm by chris

Math
NaOH v= 30mL, M=0.10 HCI v= 25.0 mL, M=? You need exactly the same number of moles (or molecules) of NaOH as of HCl .1 * 30 = M * 25 M of HCl = .1 (30/25) = .12
Monday, March 8, 2010 at 11:11am by Damon

Chemistry
Moles NaOH used = M x L. From your data, that is 0.0989 M x 0.01434 L = 0.001418 moles NaOH. You have 1.0 x (25.00 mL/100 mL) = 0.25 g of the acid used in the titration but I don't see the data to show that it is mono, di, or triprotic acid.
Monday, October 4, 2010 at 10:17pm by DrBob222

chemistry
15.00 ml of NaOH is obtained in an Erlenmeyer flask. Before titrating with .2500 M HCl, the initial volume of acid in a buret is read as 2.08 ml. At the endpoint the final volume of acid in the buret is 25.20 ml. What was the concentration of NaOH?
Wednesday, November 28, 2012 at 7:32am by lisa

Chemistry-102
what is the ph solution obtained by mixing 25.00 ml of 0.250 ml HCl and 25.00ml of 0.125M NaOH
Wednesday, March 23, 2011 at 10:58pm by Christian

Chemistry
Yes, the ratio is 1:1 because of the equation. HCl + NaOH ==> NaCl + H2O It means 0.1 mol HCl and 0.1 mol NaOH react exactly; also it means that the moles NaOH must equal the moles of HCl at the equivalence point. So if you started with 30 mL of 0.1 HCl, you started ...
Monday, May 17, 2010 at 12:56pm by DrBob222

Chemistry
HF + NaOH =NaF + H2O What will be the pH of a solution produced when 40 mL of 0.25 M HF is titrated with 80 mL of 0.125M NaOH? Ka for HF is 3.5*10^-4
Saturday, April 21, 2012 at 8:25pm by Kristen

Chemistry
I have troubles with a titration of NaOH and HCl. The volume of the HCl is 30 ml and its concentration is 0.1 mol per liter The volume of the NaOH is 25 ml and the concentration is to be worked out. I understand that the reaction ratio for HCl to neutralize NaOH is 1:1. But ...
Monday, May 17, 2010 at 12:56pm by Jazz

Chemistry
Thank you very much. The molarity of the NaOH solution is 0.162. So I did the following: 25.00 mL NaOH*(1 L / 1000 mL)*(0.160 mol NaOH / 1 L NaOH)*(1 mol CH3COOH / 1 mol NaOH) = 0.004 mol CH3COOH (0.004 mol CH3COOH / 5.00 mL) * (1000 mL / 1 L) = 0.8 M CH3COOH Now I'm ...
Wednesday, March 4, 2009 at 10:23pm by Jake

chemistry
HC2H3O2 + NaOH ==> NaC2H3O2 + H2O initial: HC2H3O2 = M x mL = 0.1 x 8.00 mL = 0.8 mmoles. NaOH = M x mL = 0.1 x 12.90 mL = 1.29 mmoles. NaC2H3O2 = 0 final: You have more NaOH than HC2H3O2; therefore, all of the HC2H3O2 will be used and some of the NaOH will remain. ...
Monday, May 3, 2010 at 2:00pm 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 desperately
You are confused because you have too much information piled in together. You need to think this through. And you don't say what unit the concentration should be in. M(NaOH) x L(NaOH) = mols NaOH. mols NaOH = mols acetic acid since the equation is 1 mol NaOH to 1 mol ...
Tuesday, December 16, 2008 at 12:10am by DrBob222

Science (chemistry)
I think the problem is to be made up to meet that criteria. For example, if we start with 25 mL of 0.1M HCl we start with 2.5 millimols. For the end point to be at 26.4 cc we need to use 0.0947 M NaOH. Then zero mL NaOH, (H^+) = 0.1 M and pH = 1. End point is 25*0.1/26.4*0.947...
Friday, March 15, 2013 at 6:19am by DrBob222

college chemistry
You are given solutions of HCl and NaOH and must determine their concentrations. You use 38 mL of NaOH to titrate 100 mL of HCl and 23.2 mL of NaOH to titrate 50 mL of 0.0782 M H2SO4. Find the unknown concentrations.
Sunday, October 3, 2010 at 3:06pm by KELLY

Chem Question
What is the molarity of a NCL solution if 37.50 mL of the acid is needed to neutralize 25.00 mL of 0.0725 M NaOH? I'm not quite sure how to go about doing this one. And also this one thing. The following technical errors were committed in standardizing the NaOH. What will ...
Thursday, March 8, 2007 at 10:25am by Angie

Chemistry
Calculate the number of mol of NaOH in 25.5 mL of 1.08 M of NaOH.
Sunday, March 9, 2008 at 9:29pm by Allie

Chemistry
I've been working on this problem for hours and cant figure it out. I know it should be simple, but I'm stuck... The question is... Describe how to prepare 1.00L of a 1.0% (w/v) NaOH from a 2.0 M NaOH. *I need to figure out how many ml of the stock solution is needed. ...
Thursday, January 10, 2013 at 6:08pm by James

chemistry
1). Adding NaOH to the buret after rinsing the buret with distilled water (essentially saying that you don't rinse the buret with NaOH)means that the NaOH is more dilute so it will take more NaOH from the buret to reach the equivalence point. So more mL means mL x M = a ...
Thursday, October 1, 2009 at 7:48pm by DrBob222

chemistry
how many milliliters of 3.25 M NaOH would be needed to make 150.0 mL of 1.50 M NaOH?
Thursday, October 7, 2010 at 6:27pm by christian

Biochemistry
millimoles NaOH = 5 mL x 0.1M = 0.5 millimoles HAc = 25 mL x 0.2M = 5 ...........NaOH + HAc ==> NaAc + H2O initial....0.5.....5.0.....0.......0 change....-0.5....-0.5...+0.5...+0.5 equil.......0.....4.5......0.5....0.5 Substitute into the Henderson-Hasselbalch equation ...
Thursday, September 8, 2011 at 11:57pm by DrBob222

Chemistry-Is this correct?
The concentration of NaOH is normally in the range of 3-6 M when they peel potatoes and soak them in a solution of NaOH , remove them and spray them off. If I have 45.7 mL of 0.500M of H2SO4 required to react completely with a 20.0 mL sample of NaOh. What is the molar ...
Thursday, May 20, 2010 at 4:20pm by Sara

Chemistry
moles H2X present = 0.1873/85.00 = ?? moles NaOH = 2 x that number. MNaOH = moles NaOH/L NaOH Solve for L NaOH and convert to mL. 41.89 mL is the correct answer.
Wednesday, May 11, 2011 at 9:06pm by DrBob222

Chemistry
The density of the solution is 1.17 g/mL. How much does a liter weigh? 1.17 g/mL x 1000 mL = ?? grams = mass of soln. How much of that is NaOH? 0.0357%; therefore, ?? g soln x 0.0357 = ??g NaOH.(The remainder is the mass of the water in the solution). How many mols NaOH is ...
Sunday, September 7, 2008 at 1:24pm by DrBob222

Chem
Tt is found that an average volume of 24.25 ml of approximately 0.1 M NaOH is required to reach an endpoint in a titration with 25.00ml of 0.09917 M HCl solution. What is the NaOH concentration? Make sure to give answer to four significant figures.
Friday, May 3, 2013 at 1:11am by Abenaa

Chemistry
Why does the addition of 50ml of 0.5M NaOH to 50mL of 0.6M KOH cause the pH to decrease (aka more acidic)? Because you have reduced the (OH^-) of the 0.6 M KOH by adding an equal volume of 0.5 M NaOH 50 mL x 0.6 M KOH = 30 millimols. pH = 13.78 Adding 50 mL x 0.5 M NaOH = 25 ...
Tuesday, August 21, 2007 at 7:34am by Vic

chemistry
5 mL x 0.0100M = 0.0500 millimoles NaOH. 25.0 mL x 0.01M HAc = 0.250 mmoles HAc. .............NaOH + HAc ==> NaAc + H2O initial...0.0500..0.25.....0......0 change...-0.0500..-0.0500.0.0500.0.0500 equil......0.......0.20....0.05..0.05 Substitute the equilibrium ...
Friday, October 21, 2011 at 8:40pm by DrBob222

chemistry
What is the ion in the resulting solution? 100.0 mL of 0.015 M K2CO3 and 100.0 mL of 0.0075 M BaBr2? 25 mL of 0.57 M Ni(NO3)2 and 100.0 mL of 0.150 M NaOH
Monday, October 24, 2011 at 8:25pm by dude

Chemistry
I'm a little confused as to the question. If you want percent water (not counting the solutions) it is (4/10)*100 = ? If you want to count the water in the aq NaOH too that is almost 2 mL (0.01M NaOH is only 0.4 g NaOH in a liter which will be 0.4 mg in 1 mL or 0.8 mg in 2...
Thursday, February 9, 2012 at 9:45pm by DrBob222

Chemistry
25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH. my work CH3COOH + H2O <-> H3O^+ + CH3COO^- 25mL x 0.100 mmol/ml = 2.5mmol CH3COOH 27mL x 0.100 mmol/ml = 2.7mmol NaOH ...
Wednesday, June 30, 2010 at 2:28am by Anonymous

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

Chemistry
Ok, great! Thank you. Sorry about question 1, I left out the full question. This is it: What is the molarity of a NaOH solution if 48.0 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of NaOH solution?
Sunday, September 19, 2010 at 11:32am by Anonymous

chemistry
Run # Volume NaOH used (mL) 1 4.1 2 4.3 3 2.8 Concentraion of NaOH: 0.113 M pH of acid: 2.84 the questions are: #moles of NaOH: ? initial acid concentration: ? Determine the Ka and pka of the weak acid: ? So here is what I did: Shaked for several minutes about 1 g of the acid ...
Sunday, July 15, 2012 at 7:53pm by fenerbahce

chem
1. 50.00 mL of an unknown monoprotic acid is titrated with 0.132 NaOH. It takes 25.25 mL of base to reach the equivalence point. Calculate the concentration of the acid.
Sunday, July 25, 2010 at 10:53am by Anonymous

Chemistry
A 50.0 mL sample of 0.12 M formic acid, HCOOH, a weak monoprotic acid, is titrated with 0.12 M NaOH. Calculate the pH at the following points in the titration. Ka of HCOOH = 1.8 multiplied by 10-4. What is the pH when 50.0 mL NaOH is added 60.0 mL NaOH is added 70.0 mL NaOH is...
Sunday, April 3, 2011 at 5:13pm by George

Introductory Chemistry
In a titration 25.1ml of a monoprotic weak acid is neutralized with 25.1 ml of .502M NaOH(aq). Calculate the Ratio of [A-] to [HA] after 12.55 of the .502 M NaOH has been added to the initial amount of HA. Anyone have any idea what's going on. I got as far as to find the ...
Wednesday, February 4, 2009 at 8:12pm by Nina

chemistry
calculate the morality of THE H2SO4 solution for each of the three runs. do the three values agree closely together with each other. Part A. Titration 1 volume of unknown acid =20ml initial buret reading of NaOH =0.5 ml final buret reading of NaOH=10.0ml Titartion 2 volume of ...
Wednesday, October 17, 2012 at 1:33pm by ALECIA

chem--please help me!!
when a 25.0 mL sample of an unknown acid was titrated with a 0.100 M NaOH solution. Determine Ka for the unknown acid. Volume NaOH = 12.5 mL; pH=3.80 please help me explain in details!!!
Saturday, October 30, 2010 at 2:29pm by Sue

Chemistry
HCl + NaOH ==> NaCl + H2O How many milliequivalents(m.e.) HCl were used? m.e. = mL x N A m.e. of HCl = m.e. NaOH = m.e. of anything. Then N = m.e. NaOH/mL NaOH.
Saturday, June 25, 2011 at 4:36am by DrBob222

Chemistry
The original post says 25 mL of 2.00 M NaOH = 25 x 2 = 50 millimoles or 0.05 moles NaOH. But your solution says 0.2 x 0.025 = ??. Something is awry. Is it 2.00 M or 0.200 M? I think that's the trouble spot.
Monday, August 10, 2009 at 12:12pm by DrBob222

chemistry
CH3COOH we will call HAc 100 mL x 0.017 = 1.7 millimoles HAc 10 mL x 0.026 = 0.25 mmoles NaOH. ...........HAc + NaOH ==> NaAx + H2O initial....1.7....25.......0......0 change....-.25..-0.25......0.25 equil.....1.45.....0........0.25 concn acid = 1.45 mmoles/(100+10)mL...
Wednesday, October 12, 2011 at 11:20pm by DrBob222

Chemistry
One commercial method used to peel potatoes is to soak them in a solution of NaOH for a short time, remove them from the NaOH, and spray off the peel. The concentration of NaOH is normally in the range of 3 to 6 M. The NaOH is analyzed periodically. In one such analysis, 45.7 ...
Wednesday, January 12, 2011 at 1:06am by Chin

chem.
Thanks. I can help with this. And the experiment part of my question meant that if an experiment I would need to know what you were doing. As you have written it the problem now makes sense. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O First, how many moles NaOH do you have in the ...
Saturday, March 19, 2011 at 4:10pm by DrBob222

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