Question

An aqueous solution contains dissolved and . The concentration of is 0.50 M and pH is 4.20. Calculate the concentration of in this buffer solution.

Answer 1

Here is the correct format for the question.

An aqueous solution contains dissolved $\mathbf{C_6H_5NH_3Cl}}$ and $\mathbf{C_6H_5NH_2}$. The concentration of $\mathbf{C_6H_5NH_2}$ is 0.50 M and pH is 4.20. Calculate the concentration of $\mathbf{C_6H_5NH_3^+}$ in this buffer solution. $\mathbf{K_b = 3.8 \times 10^{-10}}$

Answer:

$\mathbf{C_6H_5NH_3^+ = 1.1995 M }$

Explanation:

From the above information:

$pKb \ of \ C _6H_5NH_2 = -log(3.8 \times 10^{-10})$

$pKb \ of \ C _6H_5NH_2 = 9.42$

pH = 4.20

pOH = 14 - pH

pOH = 14 - 4.20 = 9.8

According to Henderson Hasselbalch equation

$pOH = pKb + log \dfrac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]}$

$9.8 = 9.42 + log \dfrac{[C_6H_5NH_3^+]}{(0.50)}$

$9.8-9.42 = log \dfrac{[C_6H_5NH_3^+]}{(0.50)}$

$0.38 = log \dfrac{[C_6H_5NH_3^+]}{(0.50)}$

$10^{0.38} = \dfrac{[C_6H_5NH_3^+]}{(0.50)}$

$2.399 = \dfrac{[C_6H_5NH_3^+]}{(0.50)}$

$C_6H_5NH_3^+ = 2.399 \times(0.50)$

$\mathbf{C_6H_5NH_3^+ = 1.1995 M }$