Question

Given, $\text{pH}$ of $\text{0}\text{.1 M NaHC}{{\text{O}}_{\text{3}}}$ if ${{\text{K}}_{\text{1}}}\text{ = 4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-7}}}\text{ , }{{\text{K}}_{\text{2}}}\text{ = 4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-11}}}$ .
A) $8.34$
B) $9.6$
C) $8.2$
D) $9.9$

Answer 1

The $\text{pH}$ for the polyprotic acid is equal to half the sum of the $\text{pka}$ values. The $\text{pka}$ values are calculated by taking the negative logarithm of dissociation constant.
$\text{pka=-log (ka)}$.

Complete answer:
Sodium bicarbonate $\text{NaHC}{{\text{O}}_{\text{3}}}$ is a sodium salt of carbonic acid${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$. Carbonic acid is a weak acid. It does not undergo complete dissociation. The carbonic acid is polyprotic. The carbonic acid dissociates to give the two protons in solution. The dissociation of carbonic acid is as follows:
The first dissociation of carbonic acid as follows,
${{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\rightleftharpoons {{\text{H}}^{\text{+}}}\text{+HCO}_{\text{3}}^{\text{-}}\text{ }{{\text{K}}_{{{\text{a}}_{\text{1}}}}}=\dfrac{\left[ \text{HCO}_{\text{3}}^{\text{-}} \right]\left[ {{\text{H}}^{\text{+}}} \right]}{\left[ {{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}} \right]}$
The second dissociation constant for carbonic acid is as:
$\text{HCO}_{\text{3}}^{\text{-}}\text{ }\rightleftharpoons {{\text{H}}^{\text{+}}}\text{+ CO}_{\text{3}}^{\text{2-}}\text{ }{{\text{K}}_{{{a}_{2}}}}=\dfrac{\left[ \text{CO}_{\text{3}}^{\text{2-}} \right]\left[ {{\text{H}}^{\text{+}}} \right]}{\left[ \text{HCO}_{\text{3}}^{\text{-}} \right]}$
Here $\text{NaHC}{{\text{O}}_{\text{3}}}$ undergoes the dissociation to give:
$\text{NaHC}{{\text{O}}_{\text{3}}}\to \text{ HCO}_{3}^{-}\text{ + }{{\text{H}}^{\text{+}}}$
From here let's find out the $\text{pH}$ solution. To do so first calculate the $\text{pka}$ values for $\text{ }{{\text{K}}_{{{\text{a}}_{\text{1}}}}}$ and $\text{ }{{\text{K}}_{{{a}_{2}}}}$.
$\text{pka}$Values are calculated by taking the negative log of the dissociation constant values.
$\text{pk}{{\text{a}}_{\text{1}}}=-\log \text{ ( 4}\text{.5}\times \text{1}{{\text{0}}^{\text{-7}}})$
We have, $\text{pk}{{\text{a}}_{\text{1}}}=6.34$
Similarly, $\text{pk}{{\text{a}}_{\text{2}}}=-\log \text{ ( 4}\text{.5}\times \text{1}{{\text{0}}^{\text{-11}}})$
We have, $\text{pk}{{\text{a}}_{2}}=10.34$
We know that for polyprotic acids the $\text{pH}$is equal to half the sum of $\text{pka}$values of first and second dissociation constant.
Therefore, $\text{pH=}\dfrac{\text{1}}{\text{2}}\text{ }\left( \text{pK}{{\text{a}}_{\text{1}}}\text{ + pK}{{\text{a}}_{\text{2}}} \right)$
Let us substitute the values for $\text{pka}$values we get,
$\text{pH=}\dfrac{\text{1}}{\text{2}}\text{ }\left( \text{6}\text{.34 + 10}\text{.34} \right)$
Or $\text{pH=}\dfrac{\text{1}}{\text{2}}\text{ }\left( 16.68 \right)$.We get,
Or $\text{pH=8}\text{.34}$
Thus, the pH for the $\text{0}\text{.1 M NaHC}{{\text{O}}_{\text{3}}}$solution is $8.34$ .
The sodium bicarbonate is a salt of carbonic acid. Even though it liberates hydroxide ions. Its pH can be calculated using the dissociation constant of its corresponding carbonic acid.

Hence, (A) is the correct option.

Note:
There is one more method to determine the $\text{pH}$ solution. This method can be said as the condensed form of the above-explained method.
For polyprotic acid having more than one dissociation constant, the concentration of proton is given by,
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=}\sqrt{{{\text{K}}_{\text{1}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{K}}_{\text{2}}}}$
Let's substitute the values we get,
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=}\sqrt{\text{(4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-7}}}\text{)(4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-11}}}\text{)}}=\sqrt{2.025\times {{10}^{-17}}}$
Thus, $\left[ {{\text{H}}^{\text{+}}} \right]\text{=4}\text{.5}\times \text{1}{{\text{0}}^{\text{-9}}}$’
We get, $\text{pH= -log }\left( 4.5\times {{10}^{-9}} \right)=8.34$