A 1 liter solution contains 0.368 M hydrocyanic acid and 0.491 M sodium cyanide. Addition of 0.092 moles of barium hydroxide will:

Answer 1

A water solution of HCN and NaCN functions as a buffer, the pH of which can be obtained from the Henderson-Hasselbach equation:

pH = pKa + log [CN-] / [HCN], pKa = 9.2

pH = 9.2 + log 0.491/0.368 = 9.3.

The addition of 0.092 moles of Ba(OH)2 will react with the HCN as follows:

2HCN + Ba(OH)2 -à Ba+ 2(CN-) +2H2O.

So, the reaction will consume 2 x 0.092 moles of HCN and produce 2 x 0.092 moles of CN-. This changes the concentration of the buffer to

[CN-] = 2 x 0.092 + 0.491 = 0.675M and [HCN] = 0.368 – 2 x 0.092 =0.184M.

The pH will now be 9.2 + log 0.675/0.184 = 9.8. The solution becomes more alkaline.