Question

Calculate the $pH$ of a buffer solution prepared by dissolving $30g$ of $N{a_2}C{O_3}$ in $500mL$ of an aqueous solution containing $150mL$ of $1M$ $HCl$ . (${K_a}$ for $HCO_3^ -$ = $5.63 \times {10^{ - 11}}$ )
A. $8.197$
B. $9.197$
C. $10.437$
D. $11.197$

Answer 1

A buffer solution is an aqueous solution that consists of a mixture of a weak acid and its conjugate base, or vice versa. Its $pH$ changes up to a very small margin, when a very small amount of a strong acid or a base is added to it. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications.

Complete step by step answer:
The reaction of the hydrochloric acid with the sodium carbonate to neutralize the corresponding base is as follows:
$N{a_2}C{O_3} + 2HCl \to 2NaCl + {H_2}C{O_3}$
The number of moles is determined by the following formula:
$n = \dfrac{w}{{{M_w}}}$
Where, $n =$ number of moles
$w =$ weight of the given solute
${M_w} =$ molecular weight of the given solute
Now, applying the above formula to determine the number of moles of sodium carbonate, we have:
${n_{N{a_2}C{O_3}}} = \dfrac{w}{{{M_w}}} = \dfrac{{30}}{{106}} = 0.28$
The number of moles is also determined by the following formula:
$n = M \times V$
Where, $n =$number of moles
$M =$ molarity of the solution
$V =$ volume of the solution
Now, applying the above formula to determine the number of moles of hydrochloric acid, we have:
${n_{HCl}} = 1 \times \dfrac{{150}}{{1000}} = 0.15$
Thus, after neutralization, the remaining moles of $N{a_2}C{O_3} = 0.28 - 0.15 = 0.13$
Thus, concentration of $N{a_2}C{O_3} = \dfrac{n}{V} = \dfrac{{0.13}}{{(0.15 + 0.5)L}}M$
The concentration of $0.15moles$ of the salt formed after neutralization = $\dfrac{n}{V} = \dfrac{{0.15}}{{(0.15 + 0.5)L}}M$
The $pH$ of the buffer solution = $pH = p{K_a} + \log \dfrac{{[salt]}}{{[base]}}$
Substituting the value of the concentration of the salt and base in the above equation, we have:
$pH = p{K_a} + \log \dfrac{{[0.15]}}{{[0.13]}}$ ….(i)
As the value of ${K_a}$ for $HCO_3^ -$ = $5.63 \times {10^{ - 11}}$
Hence, $p{K_a} = - \log (5.63 \times {10^{ - 11}}) = 11 - 0.68 = 10.32$
Substituting the value of $p{K_a}$ in equation (i), we have:
$pH = 10.32 + \log \dfrac{{0.15}}{{0.13}} = 10.32 + 0.117 = 10.437$

Thus, the correct option is C. $10.437$.

Note: The $p{K_a}$ value is one of the various methods that is used to indicate the strength of an acid. $p{K_a}$ is defined as the negative log of the acid dissociation constant or ${K_a}$ value. A lower $p{K_a}$value indicates that the acid is strong in nature. This means that the lower the value of $p{K_a}$ , the more of the acid dissociates in the water.