I only need help with the last part of the problem please!! Map Sapling Learning
ID: 490721 • Letter: I
Question
I only need help with the last part of the problem please!!
Map Sapling Learning this question hasbeencustomized by Pamela Mills at city University of New York CUNY, Lehman Coll The reaction of the weak acid HCN with the strong base KOH is HCN(aq) KoHlag) -e HoHi+ KCN (aq) To compute the pH of the resulting solution if 57mL of 0.79M HCN is mixed with 3.0 x 10 mL of 0.38 KOH we need to start with the stoichiometry. Let's do just the stoich in steps: Number Number How many moles of acid? 0.04503 How many moles of base 0.0114 What is the limiting reactant? KOH Number 0.03363 How many moles of the excess reactant after reaction? Number What is the concentration of the excess reactant after reaction? 0.3865 Number What is the concentration of the pH active product after reaction? 8.74Explanation / Answer
HCN + KOH ------------------------> KCN + H2O
57x0.79 30x 0.38 0 0 initial mmol
=45.03 =11.04
Thus the limiting reagent is KOH
33.63 0 11.04 - after reaction
The concentration of excess reagent after reaction = number of mmol/ total volume
= 33.63 /(57+30)
= 0.3865
the concentration of pH active product(KCN) after reaction = 11.04/87
= 0.1268
[KCN is pH active as it is a salt of strong base and weak acid]