30.0 mL of 0.20 M HCOOH is mixed with 30.0 mL of 0.20 M NaOH at 298 K. K_a(HCOOH) = 1.8x10^-4. Use pH = 7 + tfrac12(pK_a + log c). The pH of the mixture is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia014

30.0 mL of 0.20 M $HCOOH$ is mixed with 30.0 mL of 0.20 M NaOH at 298 K. $K_{a} (HCOOH) = 1.8 \times 1 0^{- 4}$ . Use $pH = 7 + \frac{1}{2} (p K_{a} + lo g c)$ . The pH of the mixture is

6.0 mmol of each react completely to give 6.0 mmol of sodium formate in 60 mL, so

c = 0.10

M (the dilution must be included).

pH = 7 + \frac{1}{2} (3.745 - 1) = 8.37

. Using 0.20 M would give 8.52.