Question

Determine the volume of 0.170 M NaOH solution required to neutralize each sample of hydrochloric acid. The neutralization reaction is:
NaOH(aq) + HCl(aq)---->H2O(l)+ NaCl(aq)

Part B 55 ml of a 0.065 M HCl solution

Part C 155 ml of a 0.925 M HCl solution

Answer 1

B)

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

Here:

M(HCl)=0.065 M

M(NaOH)=0.17 M

V(HCl)=55.0 mL

According to balanced reaction:

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

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

1*0.065 M *55.0 mL = 1*0.17M *V(NaOH)

V(NaOH) = 21.0294 mL

Answer: 21.0 mL

C)

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

Here:

M(HCl)=0.925 M

M(NaOH)=0.17 M

V(HCl)=155.0 mL

According to balanced reaction:

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

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

1*0.925 M *155.0 mL = 1*0.17M *V(NaOH)

V(NaOH) = 843 mL

Answer: 843 mL