Question

40 mL of 0.1 M ammonia solution is mixed with 20 mL of 0.1 M \[HCl\]. What is the pH of the mixture?
A: 5.74
B: 9.26
C: 4.56
D: 7.06

Answer 1

Hint: The letters in pH mean power of Hydrogen. pH of a solution indicates the measure of the molarity of hydrogen ions in a solution and thus we can say it is a direct measure of the alkalinity or acidity of a solution. The numerical value of pH is the negative of power of 10 of the molarity of $H^+$ ions.

Complete step by step answer:
Ammonia (\[N{H_3}\]) reacts with hydrochloric acid (\[HCl\]) to form a salt i.e. ammonium chloride (\[N{H_4}Cl\]). The chemical reaction can be depicted as follows:
$N{H_4}OH + HCl \to N{H_4}Cl + {H_2}O$
40 mL of 0.1 M \[N{H_3}\] soln = $40 \times 0.1$= 4 milli-equivalents of \[N{H_3}\] soln
Similarly, 20 mL of 0.1 M \[HCl\]= $20 \times 0.1$= 2 milli-equivalents of \[HCl\]
As \[HCl\] is deficient, it will react completely with \[N{H_3}\].
Thus, excess of \[N{H_3}\] that remains unreacted = 4−2 = 2 mol.
And the amount of \[N{H_4}Cl\] obtained is equal to the amount of \[HCl\]reacted = 2 mol.
After reaction, the solution contains 2 mol of \[N{H_3}\] and 2 mol \[N{H_4}Cl\]. Thus, it is a buffer solution.
According to Henderson-Hasselbach equation, pH of buffer solution is calculated using the following formula:
$pH = p{K_a} - \log (\dfrac{{{b^ + }}}{{BOH}})$
or
$pH = (14 - p{K_b}) - \log (\dfrac{{{b^ + }}}{{BOH}})$
Here, $BOH$= molar concentration of weak base present in the solution, \[{b^ + }\]= molar concentration of base's anion in solution which is equal to the salt’s concentration.
In the present case, concentrations of weak base as well as its salt are same owing to their amounts being same, then:
$\begin{gathered}
  \dfrac{{{b^ + }}}{{BOH}} = 1 \\
  \therefore pH = 14 - 4.74 - \log 1 = 9.26 \\
\end{gathered} $ ($pK_b$ = 4.74, log 1 = 0)

Thus, the correct answer is Option B.

Note: There is a shortcut to this long calculation. If you already know that the salt and base are having equal molarity then pH is directly equal to $pK_a$. For example in the present case, \[N{H_3}\] and \[HCl\] are having equal molarity i.e. 0.1 M. The value of $pK_b$ is 4.74 so pKa is 9.26. That means the pH of the mixture is directly equal to 9.26.