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May 5, 2016

# Homework Help: Chemistry

Posted by Donna on Tuesday, May 13, 2008 at 5:23am.

You are carrying out the titration of 100.0 mL of 1.000M acetic acid with 1.000 M sodium hydroxide. Ka = 1.76x10^-5 of acetic acid.

(a) Calculate the initial pH of your acetic acid sample.
(b) Calculate the pH of the solution after the addition of 25.0mL of NaOH.

For (a) should I just take the -log of the given Ka value, which is 4.75?

or should I do this:
CH3COOH <--> CH3COO- + H+
1.0 0 , 0
-x +x , +x
1.0-x x , x

1.0-x can be approximated to 1.0
so, x^2/1.0=1.76x10^-5
x = 4.20x10^-3, and x = [H+]
-log [H+] = pH = 2.38

I'm not sure if should just be 2.38 or 4.75

then part B, I set it up like this:
so convert both CH3COOH and NaOH to moles
.1 CH3COOH and .025 NaOH

CH3COOH + NaOH <--> CH3COONa + H2O

I subtracted moles of NaOH from CH3COOH on the reactants side, and added moles of NaOH to CH3COONa (which starts at zero) on the products side

.1-0.025 = 0.075 of CH3COOH, and 0+0.025 = 0.025 of CH3COO-

because these are conjugates, I can use Henderson-Hasselbalch:

pH = pKa + log 0.025/0.075
pH = 4.27

Is this correct? Can the pH go down when I'm adding a strong base (even if it's a smaller amount) to a weak acid?
• Chemistry - DrBob222, Tuesday, May 13, 2008 at 1:04pm

For a, the latter procedure is correct.
For b, I came out with 4.277 which I would round to 4.28 but check my arithmetic. Yes, the pH can do down because you are neutralizing 1/4 of the acid and the extra conjugate base that's formed suppresses the ionization of acetic acid more, also. Good work.

• Chemistry - DrBob222, Tuesday, May 13, 2008 at 1:31pm

Donna, I put my thinking cap on wrong this morning. There's nothing wrong with your answers. The latter method is the one to use for the first part and your second part is done very well, although, I arrived at 4.28 instead of 4.27. My error is in the explanation of acetic acid becoming more acid. It doesn't. The pH of the acid, alone, is 2.38 in the first part and becomes 4.28 when NaOH is added to form the sodium acetate salt. The acetic acid concentration is less and the pH is greater. It is more basic in the buffer solution than in 1 M acetic acid. Sorry about that. I now see you were comparing 4.75 with 4.28 but the 4.75 was never there since that isn't the way to approach the problem.

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