Answer to Question #139701 in General Chemistry for df

Question

A student mixes 50.0 mL 0f 1.00 M Ba(OH)2 with 66.4 mL of 0.494 M H2SO4. Calculate

a) the mass of BaSO4 formed (in grams).
b) the pH of the mixed solution.

Answer 1

n(Ba(OH)2) = 1.00M x 0.05L = 0.05 moles

The amount of H2SO4 is : 0.494M x 0.0664L = 0.0328 moles

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

0.0328<--- 0.0328

The mixed solution is Ba(OH)2 with 0.05 - 0.0328 = 0.0172 mol

The concentration of mixed solution is : 0.0172:(0.05 + 0.0664) = 0.1478 M

The pH of mixed solution is 14 - -log[0.1478] = 14.83

And the mass of BaSO4 is 0.0328 x ( 137 + 32 + 16 x 4 ) = 7.64 grams.