Question

6. Calculate the following quantities: a) milliliters of 0.75 M HCl required to neutralized completely 25.0 ml of 0.15 M Ba(OH)2 b) if 30.0 ml of 0.75 M HCl solution is needed to neutralize a 45 ml solution of Ca(OH)2, what is the concentration of Ca(OH)2 in the solution?

Answer 1

a)

Balanced chemical equation is:

Ba(OH)2 + 2 HCl ---> BaCl2 + 2 H2O

Here:

M(Ba(OH)2)=0.15 M

M(HCl)=0.75 M

V(Ba(OH)2)=25.0 mL

According to balanced reaction:

2*number of mol of Ba(OH)2 =1*number of mol of HCl

2*M(Ba(OH)2)*V(Ba(OH)2) =1*M(HCl)*V(HCl)

2*0.15 M *25.0 mL = 1*0.75M *V(HCl)

V(HCl) = 10.0 mL

Answer: 10. mL

b)

Balanced chemical equation is:

2 HCl + Ca(OH)2 ---> CaCl2 + 2 H2O

Here:

M(HCl)=0.75 M

V(HCl)=30.0 mL

V(Ca(OH)2)=45.0 mL

According to balanced reaction:

1*number of mol of HCl =2*number of mol of Ca(OH)2

1*M(HCl)*V(HCl) =2*M(Ca(OH)2)*V(Ca(OH)2)

1*0.75*30.0 = 2*M(Ca(OH)2)*45.0

M(Ca(OH)2) = 0.25 M

Answer: 0.25 M