How do I solve the last part of this problem The reaction of

How do I solve the last part of this problem?

The reaction of the weak acid HCN with the strong base KOH is HCN(aq) + KOH(aq) HOH(l) + KCN(aq) need to start with the stoichiometry. Let\'s do just the stoich in steps To compute the pH of the resulting solution if 69mL of 0.49M HCN is mixed with 35mL of 0.34 KOH we Number Number How many moles of acid?0.0338 How many moles of base? 0.012 What is the limiting reactant? KOH Number 0.022 Number How many moles of the excess reactant after reaction? What is the concentration of the excess reactant after reaction?21 Number What is the concentration of the pH active product after reaction? -0.9

Solution

HCN + KOH --> H2O + KCN

KCN is the pH active component after the reaction as CN- hydrolyzes and gives off OH- which measures pH of solution.

moles acid = 0.49 M x 69 ml = 0.03381 mol

moles base = 0.34 M x 35 ml = 0.0119 mol

limiting reactant = KOH

excess reactant left after the reaction = 0.02191 mol

concentration of pH active product after reaction = 0.0119 mol/0.104 L = 0.1144 M