1.) How much 3.0 M HBr is needed to neutralize 90 mL of 0.5 M NaOH? 2.)20 mL of HCl is neutralized by adding exactly 5 mL of 2.0 M KOH. What is the concentration of the HCl?

Answer 1

(1) HBr + NaOH ==> NaBr + H2O ... balanced equation

moles NaOH present = 90 ml x 1 L/1000 ml x 0.5 mol/L = 0.045 moles NaOH

moles HBr needed = 0.045 moles NaOH x 1 mol HBr/mol NaOH = 0.045 moles HBr

Volume HBr needed: (x L)(3.0 mol/L) = 0.045 moles

x = 0.015 L = 15 mls of 3.0 M HBr required

(2) HCl + KOH ==> KCl + H2O ... balanced equation

moles KOH used = 5 ml x 1 L/1000 ml x 2.0 mol/L = 0.01 moles KOH

moles HCl present = 0.01 moles HCl (1:1 mole ratio in balanced equation)

Concentration of HCl = moles/HCl/liter = 0.01 mole/0.020 L = 0.5 moles/L = 0.5 M HCl