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.

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.

the volume increases

it takes on an electronical energy