Question

Aqueous Equilibria Buffer Problem

What will be the final pH if 0.0500 moles of H+ ions are added to one liter of
a buffer made from 0.500 M hydrocyanic acid and 0.300 M sodium
cyanide? Ka for hydrocyanic acid is 6.2 x 10-10.

Solution is 8.87

Please show all steps

Answer 1

Molarity of HCN Molarity of CN- Volume of Buffer = = = 0.500 0.300 1.00 M M L Moles of component So, Moles of HCN = = Molarity (M) * Volume (V in L) 0.500 M * 1.000 L = 0.5000 mol Moles of CN- = 0.300 M * 1.000 L = 0.3000 mol Ka of the given HCN pka = = 6.200E-10 -logi ; But pKa =- log (Ka) 6.2000E-10 ) = 9.208

Moles of acid added = 0.0500 mol When H+ is added to the buffer, it reacts with base CN- and produce HCN. So during the addition of H+ to the buffer; CN- Conc/moles will decrease and HCN moles / conc will increase as shown in the below table. Net ionic equation after addition of H+ to the Buffer CN- (aq) + H+ (aq) <-------> Initial 0.300 0.050 Change -0.050 -0.050 Equil 0.250 0.000 Henderson equation for buffer solution is pH = pka + HCN (aq) 0.500 + 0.050 0.550 log Acid ] Here the Base is CN- and Acid is HCN + pH = 9.208 0.2500 log [ 0.5500 ) log [ 0.4545 ] -0.3424 = 8.865 + = = = 9.208 9.208 8.87 + + + pH of buffer after addition of (H+)