Buffer Problem - I keep getting the wrong pH and its driving me crazy! Can someone take a look?

HC2H3O2(aq) + NaOH(aq) --> NaC2H3O2(aq) + HOH(l)

200 mL .............. 180 mL

0.100M .............. 0.100M

0.020 mol .......... 0.018mol

There will be an excess of 0.002 mol HC2H3O2 and 0.018 mol C2H3O2-, both dissolved in 380 mL of solution

You will have the following concentrations

HC2H3O2 ..... 0.002 mol / 0.380 L = 0.00526M

C2H3O2^- ..... 0.018 mol / 0.380L = 0.0474M

HC2H3O2 <==> H+ + C2H3O2^- ..Ka = 1.8x10^-5

0.00526M ..........0 ......0.0474M .... initial

-x ............ ..........+x ....+x .............. change

0.00526-x .......... x ..... 0.0474+x .. equilibrium

Ka = [H+] [C2H3O2-] / [HC2H3O2]

1.8x10^-5 = x(0.0474+x) / (0.00526-x)

x = 2.0x10^-6

[H+] = 2.0x10^-6M

pH = -log[H+] = 5.70

Looks like (E) is the answer.

==============

You can also use the Henderson-Hasselbalch equation, although I'm not a fan of it, because it is too much of a "black box" -- numbers in, answer out, without understanding the chemistry.

Nonetheless.....

pH = pKa + log([base] / [acid])

pH = -log(1.8x10^-5) + log(0.0474 / 0.00526)

pH = 5.70

What is the pH of a buffer prepared by adding 180 mL of 0.100 M NaOH to 200 mL of 0.100 M acetic acid?

Note: Ka for acetic acid = 1.8x10^-5

(A) 3.79

(B) 4.34

(C) 4.74

(D) 5.04

(E) 5.70

**I know what the correct answer is, but I keep calculating a different pH and I swear to god that my setup looks right. Thanks guys for taking a look at this problem. It should be a simple buffer problem but for some reason I can't get it and I'm beginning to wonder if the answer key is wrong. If you do decide to be gracious and take the time to work this problem out, please submit your answer and please please please explain exactly how you arrived at it. Thank you so much!

Hi Adam,

moles NaOH ---> (0.100 mol/L) (0.180 L) = 0.018 mol

moles HAc ---> (0.100 mol/L) (0.200 L) = 0.020 mol

The NaOH reacts with the HAc in a 1:1 ratio to produce sodium acetate.

moles acetic acid remaining ---> 0.02 - 0.018 = 0.002 mol

moles acetate produced ---> 0.018 mol

pH = pKa + log (base / acid)

pH = 4.752 + log (0.018 / 0.002) <--- no need to recalculate new molarities since ratio of moles is same as ratio of molarities

pH = 4.752 + 0.954 = 5.706

answer choice E

NaOH + HC2H3O2 =⇒ NaC2H3O2 + H2O

0.180 liters x 0.10 moles NaOH = 0.0180 moles

moles of acetic acid = 0.200 moles x 0.10 moles acetic acid = 0.020

0.020 moles of acetic acid + 0.0180 moles of NaOH produces 0.018 moles of acetate

and leaves 0.002 moles of acetic acid

1.8 x 10^-5 = [H+](0.018/0.380L) / (0.002/0.380L)

1.8 x 10^-5 = [H+] 0.0473 / 0.00525

[H+] = 2.0 x 10^-6

pH = -(log2.0 x 10^-6) = 5.699

pH = 5.70