both 12 1 Why is the antacid analyzed using a backtitration

both, 1-2

1. Why is the antacid analyzed using a back-titration? 2. To standardize a sodium hydroxide solution, a student weighed 0.425 g of KHCgH04 (molar mass -204.22 g/mol) in a 150-mL Erlenmeyer flask and dissolved it in 25 mL of deionized water. Then he added three drops of phenolphthalein and titrated the KHP with the solution of sodium hydroxide. Calculate the molarity of the sodium hydroxide solution if 23.64 mL of this solution was needed to neutralize the KHP. KHCsH40i(aq) + NaOH(aq) H2O(l) + KNaC8H4Odaq)

Solution

Ans. #1. Antacids are generally metal hydroxides carbonates and [Al(OH)₃, Mg(OH)₂, CaCO₃ etc.]. Due to insolubility of CaCO₃ in neutral water, it can’t be titrated with strong acid. By the time CaCO₃ completely dissolved in water, there is always a bit of excess of acid in the resultant solution. Therefore, the excess acid must also be accounted by back-titration with a standardized base.

#2. Moles of KHP taken = Mass / Molar mass

= 0.425 g / (204.22 g / mol)

                                                = 0.0020811 mol

# Balanced reaction: KHC₈H₄O₄(aq) + NaOH(aq) -------> KNaC₈H₄O₄(aq) + H₂O(l)

According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol KHP (KHC₈H₄O₄).

That is,

            Moles of NaOH consumed = Moles of KHP neutralized

            Hence, Moles of NaOH consumed = 0.0020811 mol

Now,

            Molarity of NaOH solution = Moles of NaOH / Volume of solution in liters

                                                = 0.0020811 mol / 0.02364 L

                                                = 0.0880 M