Question

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

Answer 1

Answer : The pH after 30 mL of NaOH added will be 11.96

Explanation : Given,

Concentration of acetic acid = 0.10 M

Volume of acetic acid = 25 mL = 0.025 L    (conversion used : 1 L = 1000 mL)

Concentration of NaOH = 0.10 M

Volume of NaOH = 30 mL = 0.030 L

First we have to calculate the moles of $CH_3COOH$ and $NaOH$.

$\text{Moles of }CH_3COOH=\text{Concentration of }CH_3COOH\times \text{Volume of solution}=0.10M\times 0.025L=0.0025mole$

$\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}=0.10M\times 0.030L=0.0030mole$

The balanced chemical reaction is,

                           $CH_3COOH+OH^-\rightarrow CH_3COO^-+H_2O$

Initial moles      0.0025        0.0030               0

At eqm. moles       0        (0.0030-0.0025)   0.0025

                                          = 0.0005

Now we have to calculate the hydroxide ion concentration.

$[OH^-]=\frac{\text{Moles of }OH^-}{\text{Total volume}}$

$[OH^-]=\frac{0.0005mole}{(25+30)mL}=9.09\times 10^{-6}mole/mL=9.09\times 10^{-3}M$

Now we have to calculate the hydrogen ion concentration.

$[H^+][OH^-]=K_w$

$[H^+]\times 9.09\times 10^{-3}=1.0\times 10^{-14}$

$[H^+]=1.1\times 10^{-12}M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.1\times 10^{-12})$

$pH=11.96$

Therefore, the pH after 30 mL of NaOH added will be 11.96