The pH of a 3.97x10^-2 m aqueous solution of hx if its ka is equal to 3.0x10^-3 is 1.96.
pH is defined as the concentration of hydrogen ion in the solution.
Given,
Ka = 3.0x10^-3
As we know that,
Ka =( [H+] [X-])/[HX]
Let the concentration of [H+] = [X-] = x at any time t.
At the same time, concentration of [HX] = (0.0397-x)
Ka = x^2/(0.0397-x)
3.0x10^-3 = x^2/(0.0397-x)
x^2 = 0.1191 × 10^-3
x = 1.09×10^(-2)
x = 0.0109
The concentration of [H+] = 0.0109.
As we know that,
pH = -log[H+]
pH = -log(0.0109)
pH = -(-1.96)
pH = 1.96
Thus, we calculated that the value of pH of a 3.97x10^-2 m aqueous solution of hx if its ka is equal to 3.0x10^-3 is 1.96.
learn more about pH:
https://brainly.com/question/28238674
#SPJ4
