7-B. A solution of NaOH was standardized by gravimetric titration of a known qua

Question

7-B. A solution of NaOH was standardized by gravimetric titration of a known quantity of the primary standard, potassium hydrogen phthalate CO,H cO Na 3 + NaOH CO,K CO,K Potassium hydrogen phthalate CHO K, FM 204.22 The NaOH was then used to find the concentration of an unknown solution of H2SO4: H2SO4 2NaOH Na2SO4 2H20 (a) Verify from the structure of potassium hydrogen phthalate that its formula is CsH5O4K (b) Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of NaOH solution to reach the end point detected by phenolphthalein indicator. Find the concentration of NaOH (mol NaOH/kg solution) (c) A 10.00-mL aliquot of H2SO4 solution required 57.911 g of NaOH solution to reach the phenolphthalein end point. Find the molarity of H SO4.

Explanation / Answer

Ans. #a. See picture.

#b. Given-

Molar mass of KHP = 204.22 g mol-1

Mass of KHP = 0.824 g

Now,

Moles of KHP = Mass / Molar mass = 0.824 g / (204.22 g/mol) = 0.004035 mol

# Balanced Reaction of neutralization:     

KHC8H4O4(aq) + NaOH(aq) -------> KNaC8H4O4(aq) + H2O(l)

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

So,

            Moles of NaOH consumed = Moles of KHP consumed

            Hence, Moles of NaOH consumed = 0.004035 mol

            Mass of NaOH consumed = Moles x Molar mass

                                                            = 0.004035 mol x (40.0 g/ mol)

                                                            = 0.1614 g

            Mass of solvent = Mass of solution – Mass of NaOH

= 38.314 g – 0.1614 g

= 38.1526 g

= 0.0381526 kg

Now,

Molality of NaOH solution = Moles of NaOH / Mass of solvent in kg

                                                = 0.004035 mol / 0.0381526 kg

                                                = 0.10576 mol /kg

                                                = 0.10576 m

#c. We have, [NaOH] = 0.10576 mol/ kg

It’s assumed that moles of NaOH is proportional to the mass of solution.

So, moles of NaOH in 57.911 g solution = (0.004035 mol / 38.314 g soln.) x 57.911 g

                                                = 0.00610 mol

# Balanced Reaction:            H2SO4 + 2 NaOH ---------> Na2SO4 + 2 H2O

Stoichiometry: 2 mole NaOH is neutralized by 1 mol H2SO4.

According to the stoichiometry of balanced reaction, 1 mol H2SO4 is neutralizes 2 moles NaOH.

So,

            Moles of H2SO4 in sample = ½ x Moles of NaOH consumed

                                                = ½ x 0.00610 mol

                                                = 0.00305 mol

# Molarity of H2SO4 = (Moles / Volume of solution in liters) of H2SO4

                                                = 0.00305 mol / 0.010 L

                                                = 0.305 M

                       

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