Question

A student mixes 50.0 mL of 1.00 M Ba(OH)2 with 72.8 mL of 0.375 M H2SO4.

a)Calculate the mass of BaSO4 formed.
b)Calculate the pH of the mixed solution.

Answer 1

The correct answer should be 12.72 pH of the mixed solution and 9.94g mass of solid BASO4 formed.

The molecules of Ba(OH)2 are : 1.00M x 0.05L = 0.05 ( moles )

The molecules of H2SO4 are : 0.494M x 0.0864L = 0.0426816

Ba(OH)2 + H2SO4 ----> BaSO4 + 2 H2O

0.0426816<--- 0.0426816

The mixed solution is Ba(OH)2 with 0.05 - 0.0426816 = 0.0073184

The concentration of mixed solution is : 0.0073184 : ( 0.05 + 0.0864 ) = 0.054 M

The pH of mixed solution is 14 - -log[0.054] = 14 - 1.27 = 12.73 PH

And the mass of BaSO4 is 0.0426816 x ( 137 + 32 + 16 x 4 ) = 9.94 gams.