Solution of $0.1 \,N\, NH_4OH$ and 0.1 N $NH_4Cl$ has pH 9.25. Then $pK_b$ of $NH_4OH$ :-

Question

In an acidic buffer, if the ratio of base to acid concentration is increased by 100 times, then find the increase in pH of the buffer solution.

Answer 1

To solve this question, we need to use the concept of buffer solution pH calculation. A buffer solution is made of a weak base and its conjugate acid, such as NH_4OH (a weak base) and NH_4Cl (its conjugate acid). The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation for bases, which is:

\text{pH} = \text{pK}_b + \log \frac{[\text{Base}]}{[\text{Acid}]}

Given:

pH = 9.25
Concentration of NH_4OH = 0.1 N
Concentration of NH_4Cl = 0.1 N

Since both concentrations are equal, the equation simplifies to:

\text{pH} = \text{pK}_b + \log \frac{0.1}{0.1}

\log \frac{0.1}{0.1} = \log 1 = 0

So, the equation becomes:

\text{pH} = \text{pK}_b

From the given data:

9.25 = \text{pK}_b

Thus, the \text{pK}_b of NH_4OH is actually 4.75, considering the relation between \text{pH} and \text{pKa}, since:

\text{pKa} = 14 - \text{pK}_b

Substituting the values:

9.25 = 14 - \text{pK}_b

Calculating \text{pK}_b:

\text{pK}_b = 14 - 9.25 = 4.75

Thus, the correct answer is 4.75.

This explains how the equilibrium and buffer equation lead to the calculation of \text{pK}_b. Understanding and applying buffer solutions concepts are critical for solving such problems.

Answer 2