Question

What is the pH of solution made by adding 3.9g $NaN{{H}_{2}}$ into water to make 500 mL solution ${{K}_{a(N{{H}_{3}})}} = 2\times {{10}^{-5}}$[Na = 23, n = 14, H = 1]
A) 9
B) 0.7
C) 5.3
D) 13.7

Answer 1

Hint: The answer here is based upon the calculation of total moles of sodamide present and substituting it in the pH formula for solution of strong base – weak acid salt which gives the required answer.

Complete Solution :
Let us concentrate on calculating pH for the mixture of weak base and strong acid salt.
- pH is the measure of acidity and basicity of a solution and is defined as ‘negative logarithm of concentration of the hydrogen ions present in the solution’.
Here, when weak base and strong acid are missed then we get an acidic solution.

- Therefore, pH calculation for the mixture of strong acid salt and weak base that is of sodamide involves calculating using the formula,
\[pH=p{{K}_{w}}+\dfrac{1}{2}(p{{K}_{a}})+\dfrac{1}{2}\log C\]
where, $p{{K}_{w}}$ is the dissociation constant of water which has the value = 7
$p{{K}_{a}}$ is the dissociation constant for acid.
$\log C$ is the concentration of the given solution.

From the data of molar masses of the atoms we have,
Total concentration of $NaN{{H}_{2}}$ = $\dfrac{3.9}{39}\times \dfrac{1000}{500}=0.2M$
Therefore,$[NaN{{H}_{2}}] = 0.2M$
and also given that ${{K}_{a(N{{H}_{3}})}} = 2\times {{10}^{-5}}$$\Rightarrow p{{K}_{a}} = 5-\log 2$
Now, by substituting all these values in the equation of pH, we have
\[pH=7+\dfrac{1}{2}\times \left( 5-\log 2 \right)+\dfrac{1}{2}\log 0.2\]
Therefore, pH = 9
So, the correct answer is “Option A”.

Note: Confusions may occur when the questions are given in the form of ${{K}_{a}}$ rather than $p{{K}_{a}}$. Make sure that ${{K}_{a}}$ is converted into $p{{K}_{a}}$ and not substituted directly into the equation which leads to wrong answers.