Question

pipette 50 ml of the approximately 0.004 M potassium permanganate solution to a 250 ml conical flask. Note that this time the permanganate is added to the contents of the flask by pipette rather than from the burette - the principle being that it is present in excess so that the pink colour shou ld remain strongly after the nitrite addition . Using a measuring cylinder , add 25 ml of 2 M sulphuric acid. Pipette 25 ml of the previously prepared approximately 0.0100 M “ sodium nitrite ” solution to this flask . Mix the solutions thoroughly. v. Back t itration of remaining permanganate against iron(II) ammonium sulphate The residual pink colour as a result of the remaining excess permanganate ions is titrated against the approximately 0.02 M iron (II) ammonium sulphate solution (in the burette) , whose ac curate concentration was determined in step (iii) . Titrate until the pink becomes very faint , as in ; add 1 ml of N - phenylanthranilic acid and titrate to the colour change from purple to yellow.

From the result of your titration in part (v), calculate the number of moles of permanganate ion in the solution, and hence the number of moles of permanganate that reacted with nitrite ion in step (iv). From this, calculate the number o f moles of nitrite ion in 25 ml of the original approximately 0.0100 M solution, and then the actual concentration of sodium nitrite in this solution in mol dm – 3 . Hence calculate the percentage purity of the supplied sodium nitrite

Mass of sodium nitrite used = 1.7275 g

Answer 1

Concentration of potassium permanganate solution =0.004 M

Volume of potassium permanganate solution = 50 mL = 0.05 L

Moles of potassium permanganate in the solution = ( Molarity x Volume) = ( 0.004 x 0.05 ) mol = 0.0002 mol

KMnO₄ -----> K⁺ + MnO₄^-

Moles of permanganate ion in the solution = 0.0002 mol

Volume of sodium nitrite solution = 25 mL = 0.025 L= 0.025 dm³

Concentration of sodium nitrite solution = 0.01 M

Moles of sodium nitrite in the solution = ( Molarity x Volume) = ( 0.01 M x 0.025 L) = 0.00025 mol

NaNO₂ -----> Na⁺ + NO₂^-

Moles of nitrite ion in the solution = 0.00025 mol

Concentration of sodium nitrite solution = 0.00025 mol / 0.025 dm³ = 0.01 mol / dm³

Mass of 0.00025 mol sodium nitrite = ( moles x Molar mass ) = ( 0.00025 x 68.99) g = 0.0172 g

% purity =( Mass of pure sodium nitrite / Mass of sample ) x 100% = ( 0.0172 g / 1.7275 g ) x 100% = 0.99 % = 1 %.