Question

How many milliliters of 0.888 M HBr are needed to titrate each of the following solutions to the equivalence point? (a) 23.8 mL of 0.888 M CSOH ml (b) 45.6 mL of 0.533 M ROOH mL (c) 370.0 mL of a solution that contains 39.8 g of KOH per liter mL

Answer 1

HBr is a strong monoprotic acid.

a) CsOH is a strong base. It reacts with HBr as shown

$HBr+CsOH\rightarrow CsBr+H_2O$

So 1 mol HBr neutralises 1 mol CsOH

At equivalence point

Number of moles of CsOH=Number of moles of HBr

Molarity_CsOHx volume_CsOH =Molarity _HBr x Volume _HBr

0.888 M x 23.8 mL=0.888 M x V_HBr

So V_HBr=(0.888 M x 23.8 mL)/0.888 M=23.8 mL

So the volume of HBr needed to titrate given amount of CsOH to equivalence point=23.8 mL

b) RbOH is a strong base. It reacts with HBr as shown

$HBr+RbOH\rightarrow RbBr+H_2O$

So 1 mol HBr neutralises 1 mol RbOH

At equivalence point

Number of moles of RbOH=Number of moles of HBr

Molarity_RbOHx volume_RbOH =Molarity _HBr x Volume _HBr

0.533 M x 45.6 mL=0.888 M x V_HBr

So V_HBr=(0.533 M x 45.6 mL)/0.888 M= 27.4 mL

So the volume of HBr needed to titrate given amount of RbOH to equivalence point=27.4 mL

c) KOH is a strong base. It reacts with HBr as shown

$HBr+KOH\rightarrow KBr+H_2O$

So 1 mol HBr neutralises 1 mol KOH

Molar mass of KOH=Molar mass of K+Molar mass of O+Molar mass of H=39 g/mol+16 g/mol+1 g/mol=56 g/mol

Molarity of given KOH solution=(number of moles of KOH)/volume of solution (L)

=(mass/molar mass)/volume of solution (L)

=(39.8 g/56 g)/1 L=0.711 M

At equivalence point

Number of moles of KOH=Number of moles of HBr

Molarity_KOHx volume_KOH =Molarity _HBr x Volume _HBr

0.711 M x 370.0 mL=0.888 M x V_HBr

So V_HBr=(0.711 M x 370.0 mL)/0.888 M= 296.3 lpllmL

So the volume of HBr needed to titrate given amount of KOH to equivalence point= 296.3 mL