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 $NaOH$.

$\text{Moles of }CH_3COOH=\text{Concentration of }CH_3COOH\times \text{Volume of solution}=0.35M\times 0.052L=0.018mole$

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

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.

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

$[H^+]=\frac{0.0007mole}{(52+27)mL}=8.9\times 10^{-6}mole/mL=8.9\times 10^{-3}M$

Now we have to calculate the pH.

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

$pH=-\log (8.9\times 10^{-3})$

$pH=2.05$

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