Question

1.An aqueous solution of calcium hydroxide is standardized by titration with a 0.138 M solution of hydrobromic acid.
If 18.9 mL of base are required to neutralize 28.1 mL of the acid, what is the molarity of the calcium hydroxide solution?

2.An aqueous solution of hydrobromic acid is standardized by titration with a 0.194 M solution of calcium hydroxide.
If 16.7 mL of base are required to neutralize 12.6 mL of the acid, what is the molarity of the hydrobromic acid solution?

3.An aqueous solution of calcium hydroxide is standardized by titration with a 0.139 M solution of hydrochloric acid.
If 28.8 mL of base are required to neutralize 17.9 mL of the acid, what is the molarity of the calcium hydroxide solution?

Answer 1

Ca(OH)2(aq) + 2HBr(aq) ----------> CaBr2(aq) + 2H2O(l)
1 mole        2 moles
Ca(OH)2                HBr
M1 =                   M2 = 0.138M
V1 = 18.9ml            V2 = 28.1ml
n1 = 1                 n2 = 2
       M1V1/n1   =   M2V2/n2
           M1    =   M2V2n1/n2V1
                 =   0.138*28.1*1/2*18.9   = 0.1026M
2.
Ca(OH)2(aq) + 2HBr(aq) ----------> CaBr2(aq) + 2H2O(l)
1 mole        2 moles
Ca(OH)2                HBr
M1 = 0.194M            M2 =
V1 = 16.7ml            V2 = 12.6ml
n1 = 1                 n2 = 2
       M1V1/n1   =   M2V2/n2
           M2    = M1V1n2/n1V2
                 = 0.19*16.7*2/1*12.6   = 0.504M
molarity of HBr = 0.504M
3.
Ca(OH)2(aq) + 2HCl(aq) ----------> CaCl2(aq) + 2H2O(l)
1 mole        2 moles
Ca(OH)2                HCl
M1 =                   M2 = 0.139M
V1 = 28.8ml            V2 = 17.9ml
n1 = 1                 n2 = 2
       M1V1/n1   =   M2V2/n2
            M1    =   M2V2n1/n2V1
                  =   0.139*17.9*1/2*28.8   = 0.0432M
molarity of calcium hydroxide = 0.0432M