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:
[tex]HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-[/tex]
Initial: 0.035
At eqllm: 0.035-x x x
The expression of [tex]K_a[/tex] for above equation follows:
[tex]K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}[/tex]
We are given:
[tex]K_a=3.6\times 10^{-4}[/tex]
Putting values in above expression, we get:
[tex]3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033[/tex]
Neglecting the value of x = -0.0037 because concentration cannot be negative
So, concentration of [tex]H^+[/tex] = x = 0.0033 M
[tex]pH=-\log[H^+][/tex]
We are given:
[tex]H^+[/tex] = 0.0033 M
Putting values in above equation, we get:
[tex]pH=-\log(0.0033)\\\\pH=2.48[/tex]
Hence, the pH of 0.035 M aqueous aspirin is 2.48
