Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0141 M solution. The pH of the resulting solution is 2.31. Calculate the Ka for the acid. Number

Answer 1

To calculate the value of the concentration of hydrogen ion, the formula used is as follows pH =-log[H+] Now, substitute the value of pH as 2.31 in the equation 2.31-log[H] [T] = 0.0049 M HA(aq) A(a)H(aq) C0 I(M) 0.0141 C(M) -x E (M) (0.0141-x x C0 Value of x is same as the value of hydrogen ion that is 0.0049 M Therefore the equilibrium concentration of the acid (HA) = (0.0141-0.0049)M -0.0092 M

The expression for the dissociation constant of the acid is shown below. K,-[HTA] HA] (0.0049)2 0.0092 K, = 2.6×10-3 Therefore, the K, for the acid is 2.6x10-3