Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0138 M solution. The pH of the resulting solution is 2.56. Calculate the Ka for the acid.

Answer 1

Answer – Given, [HA] = 0.0138 M , pH = 2.56

We know the

pH = -log [H⁺]

so, [H⁺] = 10^-pH

             = 10^-2.56

            = 0.00275 M

We know ate equilibrium,

X = [H₃O⁺] = [A^-] = 0.00275

[HA] = 0.0138 – 0.00275

        = 0.0110 M

So, Ka = [H₃O⁺] [A^-] /[HA]

           = 0.00275*0.0275 / 0.0110 M

            = 6.87*10^-4