Question

Acetylsalicylic acid (aspirin), HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of 0.035 M aqueous aspirin at body temperature (Ka at 37°C = 3.6 × 10−4.)

Answer 1

The pH of 0.035 M aqueous aspirin is 2.48

Explanation:

We are given:

Concentration of aspirin = 0.035 M

The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

               $HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-$

Initial:         0.035

At eqllm:    0.035-x        x         x

The expression of $K_a$ for above equation follows:

$K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}$

We are given:

$K_a=3.6\times 10^{-4}$

Putting values in above expression, we get:

$3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033$

Neglecting the value of x = -0.0037 because concentration cannot be negative

So, concentration of $H^+$ = x = 0.0033 M

• To calculate the pH of the solution, we use the equation:

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

We are given:

$H^+$ = 0.0033 M

Putting values in above equation, we get:

$pH=-\log(0.0033)\\\\pH=2.48$

Hence, the pH of 0.035 M aqueous aspirin is 2.48