Question

A solid sample of zinc hydroxide is added to 0.310 L of 0.400 M hydrobromic acid. The solution that remains is still acidic. It is then titrated with 0.400 M sodium hydroxide solution, and it takes 96.5 mL of the sodium hydroxide solution to reach the equivalence point.

What mass of zinc hydroxide was added to the hydrobromic acid solution?

Answer 1

Given procedure can be summarized in the form of reactions as shown below.

PART 1 : Zn(OH) ₂ ( X g) + 2 HBr (excess) $\rightarrow$ ZnBr ₂ + 2 H₂O

PART 2 : HBr ( Remained non reacted in Part 1) + NaOH $\rightarrow$ NaBr + H₂O

First calculate initial moles of HBr taken for reaction.

We have, [ HBr] = No. of moles of HBr / volume of solution in L

$\therefore$ No. of moles of HBr = [ HBr ] $\times$ volume of solution in L

No. of moles of HBr = 0.400 mol / L $\times$ 0.310 L

No. of moles of HBr = 0.124 mol

Consider reaction, HBr + NaOH $\rightarrow$ NaBr + H₂O

From reaction, 1 mol NaOH $\equiv$ 1 mole HBr

$\therefore$ No. of moles of NaOH = [ NaOH ] $\times$ volume of solution in L = 0.400 mol / L $\times$ 0.0965 L = 0.0386 mol

$\therefore$ No. of moles of NaOH = No. of moles of HBr = 0.0386 mol

From above calculated values , we can calculate moles of HBr consumed in the reaction with zinc hydroxide.

Moles of HBr consumed in the reaction with Zn(OH) ₂ = Initial moles of HBr taken for reaction - no. of moles of HBr remained after reaction with Zn(OH) ₂

Moles of HBr consumed in the reaction with Zn(OH) ₂ = 0.124 mol - 0.0386 mol = 0.0854 mol

Now , consider reaction Zn(OH) ₂ + 2 HBr $\rightarrow$ ZnBr ₂ + 2 H₂O

From reaction, Stoichiometric ratio = No. of moles of acid / no. of moles of base = 2 /1 = 2

We have correlation, No. of moles of acid = No. of moles of base $\times$ Stoichiometric ratio

$\therefore$ No. of moles of base = No. of moles of acid / Stoichiometric ratio = 0.0854 mol / 2 = 0.0427 mol

Hence, 0.0427 mole Zinc hydroxide added to the HBr solution.

$\therefore$ Mass of Zinc hydroxide added to the HBr solution. = moles of Zn(OH) ₂$\times$ Molar mass of Zn(OH) ₂

$\therefore$ Mass of Zinc hydroxide added to the HBr solution. = 0.0427 mol $\times$ 99.42 g /mol = 4.245 g

ANSWER : Mass of Zinc hydroxide added to the HBr solution. = 4.245 g