6. Consider the following reaction: HCI (aq) NaOH (a)H20(NaCI (aq) a. If HCl is

Question

6. Consider the following reaction: HCI (aq) NaOH (a)H20(NaCI (aq) a. If HCl is the analyte and NaOH is the titrant, then sketch the titration curve for this reaction. Be sure to label the axes appropriately, and indicate the pH at the equivalence point b. 50.00 mL of HCl is titrated with 0.1752 M NaOH, and it requires 67.81 mL of NaOH to reach the equivalence point. What was the original molar concentration of HCI? c. Consider again the same titration experiment described in part b. What is the pH of the solution after 19.50 mL NaOH has been added? d. Consider again the same titration experiment described in part b. What is the pH of the solution after 84.12 mL NaOH has been added?

Explanation / Answer

Question b

Concentration of NaOH = 0.1752 M

Volume of NaOH = 67.81 ml

Moles of NaOH = 67.81 x 0.1752 /1000 = 0.01188 Moles

Moles of HCl = 0.01188 Moles

Concentration of HCl = 0.01188 x 1000 / 50 = 0.2376  M

Question C

Moles of HCl present = 0.01188 Moles  

Moles of NaOH added = 0.1752 x 19.5 / 1000 = 0.0034164 Moles

Moles of HCl present after reaction = 0.01188 - 0.0034164 = 0.00846 Moles

VOlume of solution = 50 + 19.5 = 69.5 ml

Concentration of HCl = 0.00846 x 1000 / 69.5 = 0.1217 M

pH of the solution = - log 0.1217 = 0.91

Question d

Moles of HCl present = 0.01188 Moles  

Moles of NaOH added = 0.1752 x 84.12 / 1000 = 0.014737 Moles

Moles of NaOH present after reaction = 0.014737 - 0.01188 = 0.002857 Moles

VOlume of solution = 50 + 84.12 = 134.12 ml

Concentration of HCl = 0.00846 x 1000 / 69.5 = 0.1217 M

pOH of the solution = - log 0.0213 = 1.67

pH of the solution = 14-1.67 = 12.32

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