Question

Calculate the following:

a. The pH of a 500.0 mL buffer solution containing 0.75 M HCN (Ka = 6.2 x 10^-10) and 0.55 M NaCN

b. The pH of the above buffer after the addition of 100.0 mL of 1.0 M NaOH.

c. The pH of the buffer if 100.0 mL of 1.0 M HCl was added to the solution in part a.

Answer 1

a HCN+H₂O Ka CN + H₂O+ PH is calculated by using hasselbalch fquation ! Handeson PH = Pka + log Isalt] acid] PH = pka + [EN] [HCN] pka - -log ka = -log(6:2 x1070) = .20 [HEN] = 0.75m [Nan] or [CN'] = 0.55M. - PH = 9.2 + log 0.55m. 9.2 + log (0.733) 9.07 IPH = (6) Moles of 0.75M HCN = = 0.75m x 0.5L. 0.375 moles. moles of 500mL (0.5l) of 0.55 m (N = 0.55MX0.5L R = 0.275 moles. After the addition of HCN & NaoH 100 mL of IM NaoH - CN- + H₂O.. A Moles of acid and that of (concentration of acid) will decrease CN will increase.

Moles of Naoh added = 1M x 0.1 L = Oolmel. HON + 0.375 -0.1 U 0.275 meles NaOH 0.1 - Ool o CN- 0.275 meles. +0.1 0.375 moles. 2 600x10²L. Total volume = 500 mL + 100 mL = 600m - 0.6 L [HCN ] = 0.275 moles = 0.275 M. 0.6L 0.6 [CN- ] = 0.375 ml 0.6 L - 0.375M. 0.6 .PH = 9.2 + log (0.375 x 0.6M) 1 0.275X0.6 M. - 9.2 + 0.14 TPH = 9.34 (C) After the addition of loomL of IM Hel. CN- + H+ HCN. Concentration of HON increased and that of CN will decrease. Moles of HG added = 1M XO.IL = 0. I met te

- I IC LE CN 0.275 -0.1 0.1750 +H+ ool -0.1 meles. HON. 0.375 +0.1 0.475 . LEN-] = 0.175 meles 0.6L = 0.175M 0.6 [HENI = - 0.475 meles 0.6L = 0.475 M. 0.6 PH = 9.2 + log [ON] [HCN = 9.2 + log/0.175 x 0.6M 0.475 x 0,6M 0.475. = 9.2 + log 0.115 - 9.2 + (-0.43) - 8.77 LPH = 8.77