Question

The formate ion, (CHO₂^-), is related to the acetate ion and forms ionic salts with many metal ions. Assume that 11.874 g of M(CHO₂)₂ (where M represents the atomic symbol for a particular metal) are dissolved in water. When a solution of 0.300 M sodium sulfate is added, a white precipitate forms. The sodium sulfate solution is added until no more precipitate forms, then a few excess milliliters are added. The precipitate is filtered, washed, and dried. It has a mass of 12.115 g. The filtrate is placed aside.

A potassium permanganate solution is standardized by dissolving 1.0734 g of sodium oxalate in dilute sulfuric acid, which is then titrated with the potassium permanganate solution. The principal products of the reaction are manganese(II) ion and carbon dioxide gas. It requires 21.50 mL of the potassium permanganate solution to reach the end point, which is characterized by the first permanent, but barely perceptible, pink (purple) color of the permanganate ion.

The filtrate from the original reaction is diluted by pouring all of it into a 300-mL volumetric flask, diluting to the mark with water, then mixing thoroughly. Then 15.00 mL of this diluted solution is pipetted into a 125-mL Erlenmeyer flask, approximately 35 mL of water is added, and the solution is made basic. What volume of the standard permanganate solution will be needed to titrate this solution to the end point? The principal products of the reaction are carbonate ion and manganese(IV) oxide.

Answer 1

Step 1

11.874 g (HCOO)₂M and 12.115 g MSO₄ contain the same quantity/mass (mol, g) of M .

x= molar mass of M

Molar mass of (HCOO)₂M = (90 + x) g/mol

Molar mass of MSO₄ = (96 + x) g/mol

(HCOO)₂M + Na2SO₄ = MSO₄ + 2 HCOONa

1 mol                1 mol   (ratio)           2 mol

11.874 g /(90 + x) g/mol = 12.115 g/ (96 + x) g/mol

1139.904 + 11.874x = 1090.35 + 12.115 x

49.55 = 0.241x

x = 206 g/mol (i.e. Pb, atomic mass 207.2 , considering also the valence)

Step 2

11.874 g /(90 + 206) g/mol = 0.040 mol (HCOO)₂M

The filtrate contains 2x 0.040 mol = 0.080 mol HCOONa.

Step 3 Standardization

5Na₂C₂O₄(aq) + 2KMnO₄(aq) + 8H₂SO₄(aq) ---> 2MnSO₄(aq) + K₂SO₄(aq) +            5Na₂SO₄(aq) + 10 CO₂(g) + 8 H₂O(l)

5Na₂C₂O₄(aq) + 2KMnO₄   ……..

Molar mass Na₂C₂O₄   134 g/mol

1.0734 g / 134 g/mol = 0.00801 mol Na₂C₂O₄ react with

0.00801 x 2/5 = 0.032 mol KMnO4.

0.032 mol KMnO4/ 0.0215 L = 0.15 M is the concentration of KMnO4.

Step 4. The titration of an aliquot of the filtrate.

The filtrate contains 2x 0.040 mol = 0.080 mol HCOONa.

The aliquot contains 0.080 mol x 15 mL/300 mL = 0.0040 mol HCOONa

3HCOO^- + 2MnO₄^- …… =   2MnO₂ + 3 CO2 + ….

0.0040 mol HCOONa react with 0.0040 x 2/3 = 0.0027 mol KMnO4

0.0027 mol / 0.15 mol/L = 0.01778 L = 17.80 mL KMnO4 used in titration.