Answer:
3.336.
Step-by-step explanation:
Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).
So, concentration of excess acid = [(NV)acid - (NV)base]/V total = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = 1.18 x 10⁻³ M.
For weak acids; [H⁺] = √Ka.C = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = 4.61 x 10⁻⁴ M.
∵ pH = - log[H⁺].
∴ pH = - log(4.61 x 10⁻⁴) = 3.336.