Calculate the pH of a 0.040 M aqueous solution of sodium cyanide at 298 K. K_a(HCN) = 4.9x10^-10, K_w = 1.0x10^-14. Neglect any correction to the square-root formula. Give the answer to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3ia010

Calculate the pH of a 0.040 M aqueous solution of sodium cyanide at 298 K. $K_{a} (HCN) = 4.9 \times 1 0^{- 10}$ , $K_{w} = 1.0 \times 1 0^{- 14}$ . Neglect any correction to the square-root formula. Give the answer to two decimal places.

Numerical answer type, tolerance 0.02.

K_{h} = K_{w} / K_{a} = 2.041 e - 05

[OH X^{-}] = K_{h} c = 9.035 e - 04

M, so pOH

= 3.044

and pH

= 10.96