Question

Calculate the pH of a 0.75M aqueous solution of NaCN, Ka for HCN = 6.2E-10

Calculate the pH of a 0.75 M aqueous solution of NaCN, K, for HCN is 6.2 X 1010. CNag)+ H2)HCNa)+ OH (aq) A/700 8) 9,21 15 0)479 A) 7.00 B) 9.2 C) 11.54D) 4.79 E) 2.46

Can someone explain to me how to do this problem step by step, and also explain why the reaction equation was written the way it was??

Answer 1

NaCN is a salt of weak acid (HCN) and strong Base (naOH). It will be hydrolysis by water molecule So NaCN (aq) + H_2O (aq) rightarrow HCN (aq) + NaOH (aq) we know that PH = 7 + 1/2 (PK_a + log C) {PKa = -log Ka given Ka for H CN = 6.2 times 10^-10 C = 0.75 M PH = 7 + 1/2 [-log (6.2 times 10^-10) + log (0.75)] = 11.54 The reaction is written like this because here salt (NaCN) is hydrolysed by H_2O and gives HCN and NaoH