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* \text{Volume of solution}=0.10M* 0.025L=0.0025mole`

`\text{Moles of }NaOH=\text{Concentration of }NaOH* \text{Volume of solution}=0.10M* 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^-]=(0.0005mole)/((25+30)mL)=9.09* 10^(-6)mole/mL=9.09* 10^(-3)M`

Now we have to calculate the hydrogen ion concentration.

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

`[H^+]* 9.09* 10^(-3)=1.0* 10^(-14)`

`[H^+]=1.1* 10^(-12)M`

Now we have to calculate the pH.

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

`pH=-\log (1.1* 10^(-12))`

`pH=11.96`

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