Question

Calculate the pH of the buffer prepared by mixing $600cc$ of $0.6M$ $N{H_3}$ and $400cc$ of $0.5M$ $N{H_4}Cl$ .
( ${K_b}$ for $N{H_3} = 1.8 \times {10^5}$ )
A. $11.3$
B. $9.5$
C. $9$
D. $5$

Answer 1

$pH$ is defined as the power of negative logarithm of hydrogen ion concentration. It is used to calculate acid and bases. The formula of $pH$ is given as follows: $pH = 14 - pOH$ . $pOH$ is defined as the negative logarithm of hydroxide ions concentration.

Complete step by step answer:
Buffer solutions are defined as the mixture of a weak acid and conjugate base or weak base and conjugate acid.
There are two types of buffer solution:
Acidic buffer
Basic buffer.

Acidic buffer:
It is defined as the mixture of weak acid and conjugate base.
It is a buffer that has $pH$ below $7$ .
Basic buffer:
It is defined as the mixture of weak base and conjugate acid.
It is a buffer that has $pH$ above $7$ .
Molarity is defined as the number of moles of solute that is present in per liter of the solution.
It is given by the formula: $M = \dfrac{C}{V}$
Where,
$M =$ Molarity
$C =$ number of moles
$V =$ volume per liter

Given data:
${M_1} = 0.6M$
${C_1} = 600cc$
$\therefore {V_1} = 600 \times {10^{ - 3}}$
$\therefore {V_1} = 0.6L$
Using the formula of molarity we will find out the concentration of $N{H_3}$ .
$M = \dfrac{C}{V}$
Number of moles of $N{H_3}\left( {{C_1}} \right) = {M_1} \times {V_1}$
Substituting the values we get,
Number of moles of $N{H_3}\left( {{C_1}} \right) = 0.6 \times 0.6$
Number of moles of $N{H_3}\left( {{C_1}} \right) = 0.36$
${M_2} = 0.5M$
${V_2} = 400cc$
${V_2} = 400 \times {10^{ - 3}}$
$\therefore {V_2} = 0.4L$
Number of moles of $N{H_4}Cl\left( {{C_2}} \right) = {M_2} \times {V_2}$
Substituting the values we get,
Number of moles of $N{H_4}Cl\left( {{C_2}} \right) = 0.5 \times 0.4$
Number of moles of $N{H_4}Cl\left( {{C_2}} \right) = 0.2$
${K_b}$for $N{H_3} = 1.8 \times {10^5}$
Substituting the values we get,
$p{K_b} = - \log {K_b}$
$p{K_b} = - \log (1.8 \times {10^5})$
$p{K_b} = 4.75$
$pOH = p{K_b} + \log \dfrac{{[salt]}}{{[base]}}$
Substituting the values of $p{K_b}$ , salt and base we get,
$pOH = 4.75 + \log \dfrac{{0.2}}{{0.36}}$
$pOH = 4.75 - 0.255$
$pOH = 4.495$
Therefore,
$pH = 14 - pOH$
Substituting the value we get,
$pH = 14 - 4.495$
$pH = 9.505$

So, the correct answer is Option B.

Note: if you know the $p{K_a}$ and $p{K_{{b_{}}}}$ of the given solution , then you can prepare buffer easily with known $pH$ . Buffer solution can resist the change in $pH$ even if the small amount of an acid or alkali is added.