Question

A semi-normal solution of sodium acetate in water has $\left[ {{{\text{H}}^ + }} \right]$:
A. Less than ${10^{ - 7}}\;{\text{M}}$
B. Greater than ${10^{ - 7}}\;{\text{M}}$
C. Equal to ${10^{ - 7}}\;{\text{M}}$
D. None of these

Answer 1

Hint : Sodium acetate $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}} \right)$ is a salt of weak acid and strong base. First calculate pH of the solution by using the formula of pH for weak acid and strong base. After calculating pH, the concentration of hydrogen ions,$\left[ {{{\text{H}}^ + }} \right]$ can be obtained by using pH formula.

Formulae used:
$N = \dfrac{{{\text{Gram}}\;{\text{equivalent}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume}}\;{\text{of}}\;{\text{solution}}\;{\text{in}}\;{\text{litre}}}}$
${\text{pH}} = \dfrac{1}{2}\left( {{\text{14}} + {\text{p}}{{\text{k}}_{\text{a}}} + \log C} \right)$
${\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]$

Complete step by step solution :
Normality is a concentration term. It is defined as the number of gram equivalents of solute available in one litre of solution.
$N = \dfrac{{{\text{Gram}}\;{\text{equivalents}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume}}\;{\text{of}}\;{\text{solution}}\;{\text{in}}\;{\text{litre}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {\text{I}} \right)$
When half gram equivalent of any particular solute undergoes dissolution in 1 litre of the solution, this is referred to as semi-normal solution. Thus for semi-normal solution, equation (I) can be written as:
$N = \dfrac{{\text{1}}}{{\text{2}}}$
This means the concentration of the solution is 0.5.
Since we know that sodium acetate is a salt of weak acid, i.e., acetic acid and strong base, i.e., sodium hydroxide as shown below.
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {\text{NaOH}}$
Thus for salt of weak acid and strong base, pH can be calculated by using the following formula.
${\text{pH}} = \dfrac{1}{2}\left( {{\text{14}} + {\text{p}}{{\text{k}}_{\text{a}}} + \log C} \right) \cdot \cdot \cdot \cdot \cdot \cdot \left( {{\text{II}}} \right)$
Where,
${\text{p}}{{\text{k}}_{\text{a}}}$ defines the strength of acid.
C is the concentration of solution.
The ${\text{p}}{{\text{k}}_{\text{a}}}$ value for sodium acetate is 4.54 which is a fixed value. The concentration of the solution is 0.5. Thus on substituting these values in equation (II) to calculate pH, we get
${\text{pH}} = \dfrac{1}{2}\left( {{\text{14}} + {\text{4}}{\text{.54}} + \log \left( {0.5} \right)} \right)$
Since log 5 is equal to $- 0.301$.
Thus

{\text{pH}} = \dfrac{1}{2}\left( {{\text{14}} + {\text{4}}{\text{.54}} + \left( { - 0.301} \right)} \right) \cr = \dfrac{1}{2} \times 18.239 \cr = 9.119 \cr} $$ Thus the pH of the solution is 9.119. Since we know that pH is the negative logarithm of $${{\text{H}}^ + }$$ ions concentration. Thus in order to calculate $$\left[ {{{\text{H}}^ + }} \right]$$ (concentration of hydrogen ions), use the following formula. $${\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]$$ On substituting pH value, we get $$\displaylines{ 9.119 = - \log \left[ {{{\text{H}}^ + }} \right] \cr \left[ {{{\text{H}}^ + }} \right] = {10^{\left( { - 9.119} \right)}} \cr = 7.6 \times {10^{ - 10}}\;{\text{M}} \cr} $$ Since the calculated value of $$\left[ {{{\text{H}}^ + }} \right]$$ is $$7.6 \times {10^{ - 10}}\;{\text{M}}$$ which is less than $${10^{ - 7}}\;{\text{M}}$$. Thus the correct option is A. **Note** : Whenever any solution consists of a salt which is made from weak acid and strong base, then to calculate pH of the solution, always utilize the formula $${\text{pH}} = \dfrac{1}{2}\left( {{\text{14}} + {\text{p}}{{\text{k}}_{\text{a}}} + \log C} \right)$$. The pH range of such a solution is greater than 7.