Question

A buffer solution has equal volumes of 0.1 M NH$_3$OH and 0.01 M NH$_4$Cl. The pK$_b$ of the base is 5. The pH is

• A. 10
• B. 9
• C. 4
• D. 7

Hint:

In buffer calculations, remember the relationship pK\(_a\) + pK\(_b\) = 14 for weak acid-base pairs.

Answer 1

The Correct Option is A

In a buffer solution, the pH is given by the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \] Given that the pK\(_a\) is related to pK\(_b\) by the equation pK\(_a\) + pK\(_b\) = 14, we first find pK\(_a\) by substituting the given pK\(_b\): \[ \text{pK}_a = 14 - \text{pK}_b = 14 - 5 = 9 \] The concentrations of the base (NH\(_3\)OH) and the acid (NH\(_4\)Cl) are both 0.1 M and 0.01 M.
Thus, we apply the equation: \[ \text{pH} = 9 + \log \left( \frac{0.1}{0.01} \right) = 9 + \log(10) = 9 + 1 = 10 \]
Thus, the pH is 10.

Answer 2

To determine the pH of the buffer solution, we can apply the Henderson-Hasselbalch equation for bases, which is:

pOH = pK_b + log₁₀([Salt]/[Base])

Given values:

• The concentration of NH₃OH (Base) = 0.1 M
• The concentration of NH₄Cl (Salt) = 0.01 M
• pK_b = 5

Substitute these values into the equation:

pOH = 5 + log₁₀(0.01/0.1)

pOH = 5 + log₁₀(0.1)

pOH = 5 - 1 = 4

Since pH + pOH = 14, we can find the pH:

pH = 14 - pOH = 14 - 4 = 10

Thus, the pH of the solution is 10.