i dont know how to calculate!! Does anyone help me???

Question

The hydrogen sulphate (H2S) content of a water sample was assayed with electrolytically generated iodine. After 3.00 g of potassium iodide had been introduced to a 50.0 mL portion of the water, titration required a constant current of 0.0731A for a total of 9.20 min. The reaction was as followed:

H2S (aq) + I2 (aq) <=> S (s) + 2H+ (aq) + 2I-(aq)


Calculate the concentration of hydrogen sulphide (in mg/L) of the sample.

(given: relative atomic mass: S=32.1, H=1.0)

Answer 1

David is wrong (sorry). The KI is no use until it is converted to I2 according to the equation

2 I- = I2 + 2 e-

So you need to find out how many moles of electricity you had. The way to do this is to use

Current (A) x time (s) = amount (charge in C)

C/96.486 = charge in Faradays, where the Faraday is just 1 mol of electrons.

In this case, each H2S corresponds to one I2 and therefore to 2 Faradays of electricity.

From now on in, it is just simple arithmetic.

Principles: all these electrochemical problems are just like ordinary titrations, except you need to find the amount in C and convert it to Faradays.

Answer 2

Forget the value of the current and the time it took. They are irrelevant. What is important is the concentrations and the equation.

The first thing you have to realize is that one mole of H2S reacts to form 2 moles of iodide ions. So you find the molar concentration of the potassium iodide:

Molar mass of KI = 166g
Therefore concentration of KI in solution is: 3/166 divided by 50ml = 0.3614 mol/litre

This is also the value of the concentration of iodide ions in the equation. Therefore the concentration of H2S that reacts to form the iodide ions is:

0.3614 divided by 2 = 0.1807 mol/litre

Molar mass of H2S = 33.1g

Concentration of H2S in sample is 0.1807 by 33.1 = 5.98 mg/L

Answer: 5.98 mg/L