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chem 104: acids and bases question?
The reaction of the weak acid HCN with the strong base KOH is HCN(aq)+KOH(aq)⟶HOH(l)+KCN(aq). To compute the pH of the resulting solution if 47 mL of 0.46 M HCN is mixed with 31 mL of 0.40 KOH, we need to start with the stoichiometry.
How many moles of the excess reactant after reaction? number of moles:
What is the concentration of the excess reactant after reaction? concentration:
What is the concentration of the pH active product after reaction? concentration:
1 Answer
- hcbiochemLv 711 months agoFavorite Answer
moles HCN = 0.047 L X 0.46 mol/L = 0.0216 mol HCN
moles KOH = 0.031 L X 0.40 mol/L = 0.0124 mol KOH
after reaction, moles CN- formed = 0.0124 mol
Molarity CN- = 0.0124 mol / 0.078 L = 0.159 M
Moles HCN remaining = 0.0216 - 0.0124 = 9.2X10^-3 mol
Molarity HCN = 9.2X10^-3 mol / 0.078 L = 0.118 M
Ka = [H+][CN-]/[HCN] = 6.2X10^-10
6.2X10^-10 = [H+](0.159) / 0.118
[H+] = 4.6X10^-10
pH = 9.34
