Question

Part B

Calculate the concentration (in M) of the unknown NaOHsolution in the second case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
23.00 mL	10.44 mL	0.1311 M

Part C

Calculate the concentration (in M) of the unknown NaOHsolution in the third case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
11.00 mL	10.45 mL	0.0889 M

Express your answer using three significant figures.

Part D

Calculate the concentration (in M) of the unknown NaOHsolution in the fourth case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
26.00 mL	33.18 mL	0.1021 M

Answer 1

B)
Balanced chemical equation is:
HCl + NaOH ---> NaCl + H2O


Here:
M(HCl)=0.1311 M
V(HCl)=10.44 mL
V(NaOH)=23.0 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*0.1311*10.44 = 1*M(NaOH)*23.0
M(NaOH) = 0.0595 M

Answer: 0.0595 M

B)
Balanced chemical equation is:
HCl + NaOH ---> NaCl + H2O


Here:
M(HCl)=0.0889 M
V(HCl)=10.45 mL
V(NaOH)=11.0 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*10.45 = 1*M(NaOH)*11.0
M(NaOH) = 0.0845 M

Answer: 0.0845 M

D)
Balanced chemical equation is:
HCl + NaOH ---> NaCl + H2O


Here:
M(HCl)=0.1021 M
V(HCl)=33.18 mL
V(NaOH)=26.0 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*0.1021*33.18 = 1*M(NaOH)*26.0
M(NaOH) = 0.1303 M

Answer: 0.1303 M