Zn(OH)2 solid (molecular mass = 99.41 amu) is added to 0.500 L of 0.50 M HBr. The resultant solution which is acidic, requires 185 mL of 0.500 M NaOH for neutralization. Calculate the mass (in mg) of Zn(OH)2 that was originally added (problem 53 in Purdue test bank)

Question

Zn(OH)2 solid (molecular mass = 99.41 amu) is added to 0.500 L of 0.50 M HBr. The resultant solution which is acidic, requires 185 mL of 0.500 M NaOH for neutralization. Calculate the mass (in mg) of Zn(OH)2 that was originally added  (problem 53 in Purdue test bank)

GK · Answer

The reaction is:
Zn(OH)2 + 2HBr ---> ZnBr2 + 2H2O
This reaction shows that moles of HBr = (1/2) moles of HBr.

Moles of HBr = (liters)(moles/L) or
moles HBr = (0.500 L)(0.50 mol/L) = _____?

moles of ZnBr2 = (1/2) moles of HBr

Multiply the moles of ZnBr2 by its formula mass to get the mass of ZnBr2 in grams. Convert that to milligrams.