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Determine the pH REALLY URGENT *updated*

Submitted by brunetteshorty46 on Mon, 03/29/2010 - 22:47

given acetic acid has a Ka= 1.8 x 10^(-5).  Determine the pH at the following points in the titration of 25mL of 0.5M CH3COOH with 0.5M of NaOH solution.

CH3COOH + Na +OH ---> Na + CH3COO + H2O

1) before addition of any NaOH
2) after addition of 10.00ml of 0.5M NaOH

okay, so I believe I should use the ICE chart, but I'm thrown off by the equation being Na + OH , and the questions asking for NaOH.  And then use -log(x) to find pH.

If you would help me set up the equation for both 1 & 2, I should be able to work from there. Thank you so much


For the first part, use Ka to find [H+] ions, and then find pH
CH3COOH + aq  CH3COO-(aq) + H+(aq)
Ka = [H+][CHCOO-]/[CH3COOH] which becomes Ka = [H+]2/[CH3COOH]  as [CHCOO-] = [H+]
If initial [CHCOOH] = 0.5M and the equilibrium concentration of H+ = w, then equil. conc. [CH3COOH] = (0.5 -w)

So plug these into Ka = [H+]2/[CH3COOH] using the Ka given and find w. Then find pH

In the second part write a balanced equation for the reaction. You should find that 1 mole NaOH react with 1 mole CH3COOH so that you can find the moles of CH3COOH that did not react.
(You will find that no. of moles = volume in L x molarity will be useful)
You will know the total volume of solution so you can find the new concentration of CH3COOH by moles CH3COOH/total volume in L. Then you can find the equilibrium concentration of H+ (using Ka as above) and then pH


Thank you.
For part one I got 2.52 as pH.

Part two, I'm still really confused.
  0.01L x .500M = 0.005 moles NaOH
  0.025L x 0.500M = 0.0125moles CH3COOH
Total volume of solution = 0.035L

0.0125 moles / 0.035 L = 0.357 M

-log(0.0025) = 2.60 pH

is that correctly worked out?


0.01L x .500M = 0.005 moles NaOH
  0.025L x 0.500M = 0.0125moles CH3COOH
Total volume of solution = 0.035L

So as 1 mole acid neutralises 1 mole of NaOH and there will be 0.0125 - 0.005 = 0.0075 moles of acid unused

Molarity of acid is now 0.0075/0.035 = 0.2143M
So now use Ka as before with this value of starting concentration of acid so that the new equilibrium concentration will be 0.2143-y and the equilibrium concentration of H+ will be y
Then find pH