Question

Calculate the $pH$ of a buffer solution prepared by dissolving $30$g of $N{a}_{2}C{O}_3$ in $500$ mL of an aqueous solution containing $150$ mL of $1$M $HCl$. $K_a$ for $HC{O}_3^{-}=5.63\times {10}^{-11}[log\left(\frac{133}{150}\right)=-0.05]$.

• A. $8.197$
• B. $9.197$
• C. $10.197$
• D. $11.197$

Answer 1

Answer: (C)

$10.197$

Hydrochloric acid reacts with sodium carbonate to neutralise the base as follow.

$N{a}_{2}C{O}_3+HCl\to NaCl+H_{2}C{O}_3$

$MolesofN{a}_{2}C{O}_3=\frac{m}{M}=0.28mol$

$MolesofHCl=V\left(l\right)\times M=0.15\times 1=0.15mol$

$MolesofN{a}_{2}C{O}_{3}left=0.28-0.15=0.13mol$

$ConcentrationleftN{a}_{2}C{O}_3=\frac{0.13}{\left(0.500+0.15\right)M}$

$Concentrationofsalt=\frac{0.15}{\left(0.500+0.5\right)M}$

$pHofbuffer=p{k}_a+\log \frac{\left[salt\right]}{\left[base\right]}$

$p{K}_a=-\log \left[K_a\right]$

$=-\log \left[5.63\times {10}^{-11}\right]=10.32$

$pH=10.32+\log \frac{0.15}{0.15}=10.32+0.062$

$=10.197$

Hence, the option $\left(C\right)$ is the correct answer.