Question

An antacid tablet, with magnesium hydroxide as the active ingredient, was dissolved in 20.00 mL of 1.0686 M hydrochloric solution. After the tablet completely reacted, the excess hydrochloric acid was titrated with 10.65 mL of 1.0825 M sodium hydroxide. At the completion of the titration, the solution was dark pink. Phenolphthalein was used as the indicator in this experiment. The labeled mass of the magnesium hydroxide is 311 mg per tablet. Show all work.

a) Write the molecular, ionic and net ionic equations for the reactions in this experiment.

b) Calculate the mass of magnesium hydroxide in the antacid tablet using the experimental data

c) Calculate the % recovery [(experimental value/label value) x 100]

d) Is the percent recovery consistent with the experimental error, as indicated by the final solution being dark pink? (In other words, what does it mean experimentally when the final solution in this titration with phenolphthalein as the indicator is dark pink? What impact does this titration error have on the final results, in this case, the calculated mass of the magnesium hydroxide in the antacid tablet?)

e) 10.00 mL of the sodium hydroxide solution used in this experiment diluted to 100.00 mL and this new solution was used to determine the molar mass of a monoprotic acid, HX (where X is some anion). 0.402 g of the monoprotic acid (HX) was dissolved in 100.0 mL of DI water and the solution was titrated with 18.20 mL of the dilute sodium hydroxide solution to a phenolphthalein endpoint. What is the molar mass of the acid?

f) Are any of the reactions in this problem redox reactions? Why or why not?

Please show all work :)

Answer 1

Given one tablet of antacid.which contains Mg(OH)2

The moles of HCl used to dissolve the tablet = 20 mL x 1.0686 mol / 1000 mL

                                                                 = 2.1372 x 10^-2 moles.

The excess moles of HCl = The moles of NaOH in the titration

                                     = 10.65 mL x 1.0825 mol / 1000 mL

                                     = 1.15286 x 10^-2 moles

Therefore the moles of HCl reacted with antacid = (2.1372 - 1.15286) x 10^-2 moles

                                                                     = 9.8434 x 10^-3 moles

Consider the reaction

      Mg(OH)2 + 2 HCl MgCl2 + 2 H2O

Thus 1 mole of HCl requires 0.5 moles of Mg(OH)2 to react with.

Hence the moles of Mg(OH)2 present in the antacid tablet = 0.5 x 9.8434 x 10^-3 moles

   = 4.9217 x 10^-3 moles

We know that mass = moles x molar mass

molar mass of Mg(OH)2 = 58.3 g / mol

Therefore mass of Mg(OH)2 = 4.9217 x 10^-3 x 58.3 = 0.287 g

                                           = 287 mg per tablet.

a)

The molecular equation

               Mg(OH)2 (s) + 2 HCl (aq) MgCl2 (aq) + 2 H2O (l)

The total ionic equation

             Mg(OH)2 (s) + 2 H+ (aq) + 2 Cl- (aq) Mg 2+ (aq) + 2 Cl- (aq) + 2 H2O (l)

The net ionic equation: is obtained after cancelling the2 Cl - present on both sides

               Mg(OH)2 (s) + 2 H+ (aq) Mg 2+ (aq) + 2 H2O (l)

b) the mass of Mg(OH)2 present in the antacid tablet = 287 mg as explained above.

c) The percentage recovery = experimental value x 100 / label value

                                       = 287 x 100 / 311 = 92.3 %

d) since the end point is dark pink, it shows that more volume of NaOH has been added. The actual end point should be a pale pink colour.

Thus it will increase the amount of excess HCl and hence will lower the amount of reacted HCl.

Thus, the calculated experimental amount of Mg(OH)2 is lower.

The experiment with dark pink end point lowers the calculated amount of Mg(OH)2.

e) Given that 10 mL of the NaOH is diluted to 100 mL. It is 10 times dilution. Hence the concentration of the diluted NaOH solution is 1/10 th of the concentration of the original solution which has the concentration 1.0825 M

Thus the concentrtion of NaOH used = 0.10825 M

The volume of NaOH used = 18.2 mL

Therefore moles of NaOH used = 18.2 mL x 0.10825 moles / 1000 mL = 1.97 x 10^-3 moles

since it is a monoprotic acid which is titrated with,

the moles of HX = the moles of NaOH

= 1.97 x 10^-3 moles

molar mass = mass / mole = 0.402 / 1.97 x 10^-3 [ since the mass of HX = 0.402 g ]

= 204.1 g / mol

Thus the molar mass of the acid is 204.1 g / mol.

f)

Mg(OH)2 (s) + 2 HCl (aq) MgCl2 (aq) + 2 H2O (l)

The oxidation states of Mg, O, H, Cl remains the same on both sides. Hence there is no redox reaction.

   NaOH + HCl NaCl + H2O

The oxidatin states of Na, O, H, Cl remains the same on both sides. Hence there is no redox reaction.

NaOH + HX NaX + H2O

The oxidation states of Na, O, H, X remains the same on both sides, Hence there is no redox reaction.