# How many milliliters of 1.00M KOH must be added to neutralize the following solutions?   a) a mixture of 0.240M LiOH (25.0 mL) and 0.200M HBr (75.0 mL) b) a mixture of 0.300M HCl (45.0 mL) and 0.250M NaOH (10.0mL)

In both of these instances, the mixture reacts before we add KOH. Therefore we need to find the concentrations or amounts resulting from the initial reactions in the mixtures before we proceed with KOH

a) The reaction in the mixture before adding KOH is,

LiOH + HBr -------------> LiBr + H2O

According to stoichiometry, LiOH reacts with HBr in 1:1 ratio.

Moles of LiOH present in Mixture = (0.24/1000) * 25

= 0.006 mol.

Moles of HBr present in Mixture = (0.20/1000) * 75

= 0.015 mol.

Therefore the excess HBr present in the mixture after the reaction is,

= (0.015-0.006)

= 0.009 mol.

The reaction between this excess HBr and KOH is,

KOH + HBr ---------------------> KBr + H2O

Again, KOH reacts with HBr in 1:1 ratio.

Therefore mols of KOH needed to react with 0.009 mols of HBr is 0.009 mol.

Therefore the volume needed is v,

0.009 = (1/1000) v

v = 9 ml.

Therefore volume of KOH needed to react with mixture in (a) is 9 ml.

a) The reaction in the mixture before adding KOH is,

NaOH + HCl -------------> NaCl + H2O

According to stoichiometry, NaOH reacts with HCl in 1:1 ratio.

Moles of NaOH present in Mixture = (0.25/1000) * 10

= 0.0025 mol.

Moles of HCl present in Mixture = (0.30/1000) * 45

= 0.0135 mol.

Therefore the excess HCl present in the mixture after the reaction is,

= (0.0135-0.0025)

= 0.011 mol.

The reaction between this excess HCl and KOH is,

KOH + HCl ---------------------> KCl + H2O

Again, KOH reacts with HCl in 1:1 ratio.

Therefore mols of KOH needed to react with 0.011 mols of HCl is 0.011 mol.

Therefore the volume needed is v,

0.011 = (1/1000) v

v = 11 ml.

Therefore volume of KOH needed to react with mixture in (b) is 11 ml.

Approved by eNotes Editorial Team