Question

What mass of Zn(OH)2 was added to the HBr solution?A solid sample of Zn(OH)2 is added to 0.330 L of 0.520 M aqueous HBr.

The solution that remains is still acidic. It is then titrated with 0.520 M NaOH solution, and it takes 82.5 mL of the NaOH solution to reach the equivalence point.

Answer 1

Number of moles,n= molarity x volume in L

nHBr= 0.520M x 0.330L=0.1716 mol

n NaOH= 0.520 M x 0.0825L=0.0429 mol

NaOH+ HBr----> NaBr+ H2O

1 mole NaOH reacts with 1 mole of HBr

0.0429 moles of NaOH reacts with 0.0429 moles of HBr

So number of moles of HBr reacted with Zn(OH)2 is 0.1716 mol -0.0429 mol= 0.1287 moles.

Zn(OH)2+2HBr ----> ZnBr2+ 2H2O

2 moles of HBr reacts with 1 mole of Zn(OH)2

0.1287 moles of HBr reacts with 0.1287/2=0.06435 moles of Zn(OH)2

So mass of Zn(OH)2,m = number of moles x molar mass

m= 0.06435 mol x 99.4 (g/mol)

= 6.396 g