The reaction of the weak acid HCN with the strong base KOH is: HCN(aq) + KOH(aq)
ID: 1073926 • Letter: T
Question
The reaction of the weak acid HCN with the strong base KOH is: HCN(aq) + KOH(aq) rightarrow HOH(l) + KCN(aq) To compute the pH of the resulting solution if 57mL of 0.79 M HCN is mixed with 16mL of 0.39 KOH we need to start with the stoichiometry. Let's do just the stoich in steps: How many moles of acid? How many moles of base? What is the limiting reactant? How many moles of the excess reactant after reaction? What is the concentration of the excess reactant after reaction? What is the concentration of the pH active product after reaction?Explanation / Answer
HCN + KOH ----------> KCN + H2O
no of moles of HCN = molarity * volume in L
= 0.79*0.057 = 0.045moles
no of moles of KOH = molarity * volume in L
= 0.39*0.016 = 0.00624moles
1 mole of HCN react with 1 mole of KOH
KOH is limiting reactant
HCN is excess reactant
HCN + KOH ----------> KCN + H2O
I 0.045 0.00624 0
C -0.00624 -0.00624 0.00624
E 0.03876 0 0.00624
no of moles of excess reactatn after complete of reaction HCN = 0.045-0.00624 = 0.03876moles
molarity of HCN = no of moles/total volume in L
= 0.03876/0.073 = 0.53M
molarity of KCN = no of moles/total volume in L
= 0.00624/0.073 = 0.0855M
PH = PKa + log[KCN]/[HCN]
= 9.2 + log0.0855/0.53
= 9.2-0.7923 = 8.4077 >>>>answer
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