Question

Exercise 14.48- Enhanced - with Feedback Four solutions of unknown NAOH concentration are titrated with solutions of HCL The following table lists the volume of each unknown NAOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. HCI (M) NAOH volume HCl volume (mL) (mL) 0.2099 8.00 8.97 11 14 0 1061 21.00 10 85 12.00 0.0889 35 18 29.00 0.1021 You may want to reference (Pages 494-497) Section 14.6 while completing this problem

Answer 1

A)

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

Here:

M(HCl)=0.2099 M

V(HCl)=8.97 mL

V(NaOH)=8.0 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*0.2099*8.97 = 1*M(NaOH)*8.0

M(NaOH) = 0.2354 M

Answer: 0.235 M

B)

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

Here:

M(HCl)=0.1061 M

V(HCl)=11.14 mL

V(NaOH)=21.0 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*0.1061*11.14 = 1*M(NaOH)*21.0

M(NaOH) = 0.05628 M

Answer: 0.05628 M

C)

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

Here:

M(HCl)=0.0889 M

V(HCl)=10.85 mL

V(NaOH)=12.0 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*0.0889*10.85 = 1*M(NaOH)*12.0

M(NaOH) = 0.0804 M

Answer: 0.0804 M