Answer to Question #71392 in General Chemistry for rachel

Question

What volume of a 0.537 M barium hydroxide solution is needed to exactly neutralize 7.57 grams of KHP ?

Answer 1

CM (Ba(OH)2)=0.537 M
m(KHP)=7.57 g
V(Ba(OH)2 )- ?
Ba(OH)2 + 2KHP = Ba(KP)2 +2H2O
n(KHP)= m(KHP)/M(KHP)=7.57g / 71g/mol= 0.106mol.
n(Ba(OH)2)=n(KHP)/2=0.106mol/2= 0.053mol.
CM (Ba(OH)2)= n(Ba(OH)2)/ V(Ba(OH)2 ).
V(Ba(OH)2 )= n(Ba(OH)2)/ CM (Ba(OH)2)= 0.053mol/0.537 M= 0.098 l.