Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0121 M solution. The pH of the resulting solution is 6.49. Calculate the Ka for the acid.

Answer 1

This problem is similar to that I answered yesterday, even less difficult.

1. Write chemical equation of AH with water.
2. pH allows you to know \([H_3O^+]\) and \([A^-]\)
3. deduce \([AH]\) by subtracting from molarity of solution.

4. Use expression of Ka to compute its numerical value.

Answer 2

Ka = [H3O+][A-]/[HA]
Since pH = 6.49, [H3O+] = 10^(-6.49)
[A-] = [H3O+] = 10^(-6.49) and [HA] = 0.0121

Put this into the acidity constant expression and you should get Ka= 8.65 x 10^-12