Question

Calculate the pH of a buffer solution prepared by dissolving 30g of $N{a_2}C{O_3}$ in 500 ml of an aqueous solution containing 150 ml of$1M\;HCl$. Given $\left( {{k_a}{\text{ }}for\;HC{O^{ - 3}}\; = {\text{ }}5.63{\text{ }}x{\text{ }}{{10}^{ - 11}}} \right)$
A.9.86
B.10
C.10.2
D.10.86

Answer 1

Hint : We can use the Henderson-Hasselbalch equation to determine the pH of the buffer solution.

Complete step by step solution :
Mass of sodium carbonate, $N{a_2}C{O_3}{\text{ }} = {\text{ }}30\;gm$
Volume of $HCl{\text{ }} = {\text{ }}150{\text{ }}ml$
Molarity of $HCl$used = $1M$
Complete step by step answer:
Now,
$Molar{\text{ }}Mass{\text{ }}of\;N{a_2}C{O_3} = {\text{ }}\left( {23 \times 2} \right) + 12 + \left( {16 \times 3} \right){\text{ }} = {\text{ }}106$
No of Moles of $N{a_2}C{O_3}$ ​ = $\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}} = \dfrac{{30}}{{106}}$ ​= $0.28\;mole$
$No{\text{ }}of{\text{ }}Moles{\text{ }}of\;HCl = molarity \times Volume{\text{ }}Litre$
Therefore, No of Moles of $HCl$ = $1{\text{ }} \times {\text{ }}0.15{\text{ }} = {\text{ }}0.15{\text{ }}\;mole$
As mentioned in the question, the chemical equation for the reaction is
${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + HCl}} \Leftrightarrow {\text{NaCl + N}}{{\text{a}}_{\text{2}}}{\text{HC}}{{\text{O}}_{\text{3}}}$
${\text{N}}{{\text{a}}_{\text{2}}}{\text{HC}}{{\text{O}}_{\text{3}}}{\text{}} \to {\text{N}}{{\text{a}}^{\text{ + }}}{\text{ + HC}}{{\text{O}}^{{\text{3 - }}}}$
So, during the reaction of sodium carbonate and hydrochloric acid, 0.15 moles of Na2​CO3​ will be consumed to react with hydrochloric acid.
Buffer solution is formed due to $N{a_2}C{O_3} - NaHC{O_3}$
We know that,
Equivalent of $N{a_2}C{O_3}$ = Equivalent of $HCl$
For$HCl$, Molarity is equals to Normality,
∴ Equivalent of $HCl$ = Normality ×Volume =$1 \times 0.150 = 0.150$
Equivalent of Na2​CO3 = $\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}} = \dfrac{{30}}{{106}}$ ​=$0.283$
Therefore, Amount of $N{a_2}C{O_3}$ ​ left after neutralization $= 0.28 - 0.15{\text{ }} = {\text{ }}0.13\;mole$And, 0.150 equivalent of $N{a_2}C{O_3}$ is converted to $NaHC{O_3}$
So, Equivalent of $NaHC{O_3}$ = 0.150
According to Henderson-Hasselbalch equation,

$$pH\; = \;pKa\; + \;log\;\dfrac{{\mathop {[C{O_3}]}\nolimits^{2 - } }}{{\mathop {[HC{O_3}]\;}\nolimits^ - }}\;\;\;\; \\\ pH{\text{ }}\; = - log(5.63 \times \mathop {10}\nolimits^{ - 11} ) + log\;\dfrac{{0.133}}{{0.150}} \\\ pH{\text{ }} = ( - 0.7505 + 11) + log\;\dfrac{{0.133}}{{0.150}}\;\; \\\$$ $${pH{\text{ }} = {\text{ }}( - 0.7505 + 11){\text{ }} + {\text{ }}\left( { - 0.05} \right)} \\\ {pH{\text{ }} = 10.24 + ( - 0.05)\;} \\\ {pH\; = 10.19{\text{ }} = 10.2}$$

So, pH of the buffer solution formed is 10.2
Hence, the correct option is option ‘C’

Note : We can use it to calculate the isoelectric point of protein. (Point at which Protein neither accepts nor yields proton).it is essential to have the knowledge of molar mass calculation, equivalency calculation for given chemical species to apply Henderson-Hasselbalch equation.