Question

A. 25.0 mL of 0.212 M NaOH is neutralized by 13.6 mL of an HCl solution. The molarity of the HCl solution is

B. What is the molarity of a KOH solution if 25.0 mL neutralizes 35.0 mL of a 0.200 M HCl solution?

Answer 1

A)

we have the Balanced chemical equation as:

NaOH + HCl ---> NaCl + H2O

Here:

M(NaOH)=0.212 M

V(NaOH)=25.0 mL

V(HCl)=13.6 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HCl

1*M(NaOH)*V(NaOH) =1*M(HCl)*V(HCl)

1*0.212*25.0 = 1*M(HCl)*13.6

M(HCl) = 0.390 M

Answer: 0.390 M

B)

we have the Balanced chemical equation as:

HCl + KOH ---> KCl + H2O

Here:

M(HCl)=0.2 M

V(HCl)=35.0 mL

V(KOH)=25.0 mL

According to balanced reaction:

1*number of mol of HCl =1*number of mol of KOH

1*M(HCl)*V(HCl) =1*M(KOH)*V(KOH)

1*0.200*35.0 = 1*M(KOH)*25.0

M(KOH) = 0.280 M

Answer: 0.280 M