Question

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH.
What was the Ka value calculated by the student if thepH of the solution was 2.62?

A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87.

Calculate the Kb for ethylamine.

Answer 1

Answer:

1. $3.57\times 10^{-4}$was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$was the $K_b$ of ethylamine value calculated by the student.

Explanation:

1.

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$:

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$was the $K_a$ value calculated by the student.

2.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$:

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$was the $K_b$ of ethylamine value calculated by the student.