Exercise 3 | Buffers and Titration of Acids and Bases
General Chemistry 3 - Exercise 3
Calculate the pH of a solution originally containing 0.200 mol of cyanic acid HCNO following addition of 80.0 mL of 1.00 M NaOH.
Data: Ka (HCNO/-CNO) = 3.5 x 10-4
Reaction: HCNO + HO- → -CNO + H2O
Calculate the number of moles of -CNO formed:
nCNO = nHO = 8.00 x 10-2 x 1.00 = 8.00 x 10-2 mol
nHCNO = nHCNO-initial - nHO = 1.60 x 10-1 mol
Henderson-Hasselbalch equation can be used:
pH = pKa + log
pH = pKa + log
pH = 3.28