Question

A solid sample of Zn(OH)2 is added to 0.350L of 0.540M aqueous HBr. The solution that remains is still acidic. It is then titrated with 0.540M NaOH solution, and it takes 92.5mL of the NaOH solution to reach the equivalence point. What mass of Zn(OH)2 was added to the HBr solution?

Answer 1

moles of HBr = volume of HBr x molarity of HBr

= 0.350 L x 0.540mol/L

= 0.189 mol

Moles of NaOH = volume of NaOH x molarity of NaOH

= 92.5 mL x 1L/1000mL x 0.540 mol/L

=0.04995 mol

Moles of HBr used = 0.189 - 0.04995 = 0.139 mol

The balanced reaction

Zn(OH)2 + 2 HBr = ZnBr2 + 2 H2O

1 mol Zn(OH)2 reacts with = 2 mol HBr

From the stoichiometry of the reaction

Moles of Zn(OH)2 reacted = 0.139/2 =0.0695

mass Zn(OH)2 = moles x molecular weight

= 0.0695 mol x 99.40 g/mol

=6.90 g