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

First calculate te concentration of the permangante (KMnO4) in the reaction with the oxalate (Na2C2O4) in acid media

$16H^++5Na_2C_2O_4+2KMnO_4\rightarrow 2Mn^{2+}+10CO_2+8H_2O+10Na^++2K^+$

Calculate the moles ox oxalete

$1.0734gNa_2C_2O_4 \times \frac{1mol}{134g}=8.0104\times 10^{-3}moles Na_2C_2O_4$

Calculate the moles of permangante that reacted with the oxalate

$8.0104\times 10^{-3}moles Na_2C_2O_4 \times \frac{2molesKMnO_4}{5moles Na_2C_2O_4}=3.204\times 10^{-3}moles KMnO_4$

Calculate the concentration dividing by the volume used (in Liters)

$\frac{3.204\times 10^{-3}moles KMnO_4}{.0215L}=0.149M$

From the first part, you can infer the reaction was:

$M(CHO_2)_2+Na_2SO_4\rightarrow 2(CHO_2)^-+MSO_4_{(s)}+2Na^+$

What remained in solution was the sodium and the formate (CHO2)

You need to calculate the mass of M

Assume the molecular weight of the metal formate (FM) is M+90 and for the metal sulfate (MS) that precipitated M+96

From the stiochiometry:

$11.874gMF\frac{1molMF}{M+90g}*\frac{1molMS}{1molMF}*\frac{M+96}{1molMS}=12.115gMS$

this yields the following equation:

$11.874*(M+96)=12.115*(M+90)$

Solving form you get M=205.62g/mol

Now you can calculate the concentration of formate in the solution:

M+90=295.62g/mol

$11.874gMF*\frac{1mol}{295.62g}*\frac{2molCHO_2}{1molMF}=0.08molCHO_2$

and the reaction for the last part is the permanganate with the (CHO2) in basic media is:

$OH^-+3CHO_2^-+2MnO_4^-\rightarrow 2MnO_2+3CO_3^{2-}+2H_2O$

First you diluted the 0.08 moles in 300mL

$\frac{0.0803mol}{0.3L}=0.268M$

then 15mL, you need to calculate the moles in 15 mL:

$0.268M*0.015=4.02\times 10^{-3}molesCHO_2$

From the reaction calculate the moles of permangante needed:

$4.02\times 10^{-3}molesCHO_2*\frac{2MnO_4^-}{3molCHO_2}=2.68\times10^{-3}molMnO_4^-$

And now the volume of permangante, using the concentration previously calculated:

$\frac{2.68\times10^{-3}mol}{0.149M}=0.018L*1000=18mL$ <----- volume of permanganate