Question

A 52.0-ml volume of 0.35 m ch3cooh (ka=1.8×10−5) is titrated with 0.40 m naoh. calculate the ph after the addition of 27.0 ml of naoh. express your answer numerically.

Answer 1

Answer : The pH after 27.0 mL of NaOH added will be 2.05

Explanation : Given,

Concentration of acetic acid = 0.35 M

Volume of acetic acid = 52.0 mL = 0.052 L (conversion used : 1 L = 1000 mL)

Concentration of NaOH = 0.40 M

Volume of NaOH = 27.0 mL = 0.027 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}=0.35M* 0.052L=0.018mole`

`\text{Moles of }NaOH=\text{Concentration of }NaOH* \text{Volume of solution}=0.40M* 0.027L=0.011mole`

The balanced chemical reaction is,

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

Initial moles 0.018 0.011 0

At eqm. moles (0.018-0.011) 0 0.011

= 0.0007

Now we have to calculate the hydrogen ion concentration.

`[H^+]=\frac{\text{Moles of }H^+}{\text{Total volume}}`

`[H^+]=(0.0007mole)/((52+27)mL)=8.9* 10^(-6)mole/mL=8.9* 10^(-3)M`

Now we have to calculate the pH.

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

`pH=-\log (8.9* 10^(-3))`

`pH=2.05`

Therefore, the pH after 27.0 mL of NaOH added will be 2.05

Answer 2

CH₃COOH millimoles = 52 x 0.35 = 18.2
NaOH millimoles = 27 x 0.4 = 10.8

CH₃COOH + NaOH → CH₃COONa + H₂O
18.2 10.8 0 0 -----> Initial
- 10.8 - 10.8 + 10.8 + 10.8 -----> Changed
7.4 0 10.8 10.8 ------> after reaction
Now the mixture contain acid and salt so it forms buffer
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
pH = pKa + log [salt] / [acid]
pH = 4.74 + log [10.8] / [7.4]
pH = 4.9