<u> </u> The pH of 0.035 M aqueous aspirin is 2.48
<u>Explanation:</u>
We are given:
Concentration of aspirin = 0.035 M
The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

<u>Initial:</u> 0.035
<u>At eqllm:</u> 0.035-x x x
The expression of
for above equation follows:
![K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BC_9H_7O_4%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHC_9H_7O_4%5D%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = -0.0037 because concentration cannot be negative
So, concentration of
= x = 0.0033 M
- To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
= 0.0033 M
Putting values in above equation, we get:

Hence, the pH of 0.035 M aqueous aspirin is 2.48