20.0 mL of 0.10 M formic acid (K_a = 1.8x10^-4) is titrated with 0.10 M NaOH. Calculate the pH at the equivalence point at 298 K, using pH = 7 + tfrac12(pK_a + log c) with c the concentration of the salt at that point. Give two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3ia044

20.0 mL of 0.10 M formic acid ( $K_{a} = 1.8 \times 1 0^{- 4}$ ) is titrated with 0.10 M NaOH. Calculate the pH at the equivalence point at 298 K, using $pH = 7 + \frac{1}{2} (p K_{a} + lo g c)$ with $c$ the concentration of the salt at that point. Give two decimal places.

Numerical answer type, tolerance 0.02.

At equivalence 20.0 mL of NaOH has been added, so 2.0 mmol of formate is in 40 mL:

c = 0.050

M. pH

= 7 + \frac{1}{2} (3.745 - 1.301) = 8.22