Question

${CH}_{3}COOH(50mL,0.1M)$ is titrated against $0.1M$ $NaOH$ solution.

Calculate the $pH$ at the addition of $0mL,10mL,20mL,25mL,40mL,50mL$ of $NaOH$.

$K_a$ of ${CH}_{3}COOH$ is $2\times {10}^{-5}$.

• A. $\left(i\right)1.42,\left(ii\right)2.045,\left(iii\right)2.26,\left(iv\right)2.34,\left(v\right)2.65,\left(vi\right)4.34$
• B. $\left(i\right)1.85,\left(ii\right)2.34,\left(iii\right)2.43,\left(iv\right)2.48,\left(v\right)3.21,\left(vi\right)5.43$
• C. $\left(i\right)2.85,\left(ii\right)4.0969,\left(iii\right)4.5229,\left(iv\right)4.699,\left(v\right)5.301,\left(vi\right)8.699$
• D. $\left(i\right)3.7,\left(ii\right)4.68,\left(iii\right)4.86,\left(iv\right)4.96,\left(v\right)6.42,\left(vi\right)10.86$

Answer 1

The correct option is C $\left(i\right)2.85,\left(ii\right)4.0969,\left(iii\right)4.5229,\left(iv\right)4.699,\left(v\right)5.301,\left(vi\right)8.699$

$K_a$ of ${CH}_{3}COOH$ is $2\times {10}^{-5}$.

(ii) Addition of $10ml$ of $NaOH$:

Total volume $=50+10=60mL$

Number of millimoles of NaOH added $=10ml\times 0.1M=1$ mmol.

Number of millimoles of acetic acid present initially $=50ml\times 0.1M=5$ mmol.

After partial neutralisation, $\left[C{H}_{3}COOH\right]=\frac{5mmol-1mmol}{60mL}=0.067$M

$\left[C{H}_{3}COONa\right]=\frac{1mmol}{60mL}=0.0167$M

The pH of the solution is $pH=p{K}_a+log\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$

$pH=-log(2\times {10}^{-5})+log\frac{0.0167}{0.067}$

$pH=4.0969$

(iii) Addition of $20ml$ of $NaOH$:

Total volume $=50+20=70$$mL$

Number of millimoles of $NaOH$ added $=20ml\times 0.1M=2$ mmol.

Number of millimoles of acetic acid present initially $=50ml\times 0.1M=5$ mmol.

After partial neutralisation, $\left[C{H}_{3}COOH\right]=\frac{5mmol-2mmol}{70mL}=0.042857$ M

$\left[C{H}_{3}COONa\right]=\frac{2mmol}{70mL}=0.02857$ M

The pH of the solution is $pH=p{K}_a+log\frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$

$pH=-log(2\times {10}^{-5})+log\frac{0.02857}{0.042857}$

$pH=4.5229$