Question

If the Ka of a monoprotic weak acid is 3.3 × 10-6, what is the pH of a 0.30 M solution of this acid?

Answer 1

Ans---

let us consider a monoprotic weak acid HA , basic acid dissociates as follows:

  HA --- --> H⁺ + A^-   , using ICE rule we can find the pH

Initially 0.30 0 0

change -x +x +x

at equilibrium (0.30-x) x x

now, ka=[H+][A-] /[HA]

=> 3.3 × 10^-6 = X² /(0.30-X) [As the value of ka is very small , so x in the denominator is neglected]

=>x2 = 3.3 × 10-6 * 0.30

=>x2=0.3(3.3(10−6)

Using quadratic formula , with  a=1, b=0, c=-9.899999999999998e-7 , we get

x=0.00099

so , [H+] = 0.0009M

We know , pH =-log [H+]

pH=-log(0.0009)

pH=3.04