- #1

Eclair_de_XII

- 1,083

- 91

## Homework Statement

"A 35.0-mL sample of 0.150

*M*acetic acid (CH

_{3}COOH) os titrated with 0.150

*M*NaOH solution. Calculate the pH after the following volumes of base have been added:

**(a)**0 mL,

**(b)**17.5 mL,

**(c)**34.5 mL,

**(d)**35.0 mL,

**(f)**50.0 mL."

## Homework Equations

K

_{a}= [H

^{+}][A

^{-}]/[HA]

K

_{a}= 1.8 ⋅ 10

^{-5}

x = ½(-b ± √(b

^{2}- 4ac)

pH = -log[H

^{+}]

pH = 2.78 (from back of book)

## The Attempt at a Solution

0.15 mol/L ⋅ 1 L/1000 mL ⋅ 35.0 mL = 5.25 ⋅ 10

^{-3}mol HAc

1.8 ⋅ 10

^{-5}= (x)(x)/(5.25 ⋅ 10

^{-3}- x)

x

^{2}+ 1.8 ⋅ 10

^{-5}x - 9.45 ⋅ 10

^{-8}= 0

x = ½(-1.8 ⋅ 10

^{-5}± √3.24 ⋅ 10

^{-10}- 37.8 ⋅ 10

^{-8})

x = 2.99 ⋅ 10

^{-4}mol H

^{+}

[H

^{+}] = (2.99 ⋅ 10

^{-4}mol H

^{+})/(0.0350 L) = 8.54 ⋅ 10

^{-3}M

-log(8.54 ⋅ 10

^{-3}) = 2.069 ≠ 2.78

This is rather an unorthodox, and thus, unreliable method of finding the initial pH of the acetic acid solution. But I couldn't find any other way to do it... Could someone tell me if I'm doing this problem right?