The concentration of dissolved chlorine in tap water can be determined by adding excess potassium iodide and then using aqueous sodium thiosulphate to precipitate the resulting iodide ions. In Ontario, the aesthetic objective for chlorine in tap water is a maximum concentration of 250 mg/L. Find the minimum volume of a 0.00100 mol/L solution of potassium iodide that should be used to test a 500mL sample of water.
i would be great to show me the answer , but do it step by step so i understand better
I don't know all of the details of the procedure you use but here are the steps you need.
Cl2 + 2I^- ==> I2 + 2Cl^-
So 500 mL sample x 250 mg/L means you have 250 mg/L x 0.500 L = 125 mg Cl2 present. How many moles is that?
0.125g/molar mass Cl2 = about 0.002.
Adjust for the volume of sample you will use. Surely you won't use all 500 mL.
Then moles KI soln required is twice moles Cl2 you have in the volume taken for analysis and L KI = moles KI/M KI. Solve for L and convert to mL.
How many grams of H2 would be formed if 34 grams of carbon reacted with an unlimited amount of H2O? The reaction is:
C + H2O → CO + H2
The atomic mass of C is 12.01 g/mole. The atomic mass of H2 is 2.016 g/mole
To find the minimum volume of a 0.00100 mol/L solution of potassium iodide needed to test a 500 mL sample of water, we can use the stoichiometry of the reaction between chlorine and iodide ions.
The balanced equation for this reaction is:
2Cl₂ + 2KI → 2KCl + I₂
From the equation, we can see that it takes 2 moles of chlorine to react with 2 moles of potassium iodide to produce 1 mole of iodine.
Given that the maximum concentration of chlorine in tap water is 250 mg/L, we convert this to moles by using the molar mass of chlorine (35.45 g/mol):
250 mg/L * (1 g / 1000 mg) * (1 mol / 35.45 g) = 0.00705 mol/L
Now we can set up a ratio:
0.00705 mol/L chlorine : 1 mol iodine = x mol/L potassium iodide : 0.00100 mol/L potassium iodide
Simplifying the equation, we find:
x = (0.00705 mol/L chlorine * 0.00100 mol/L potassium iodide) / 1 mol iodine
x = 0.00000705 L potassium iodide needed to react with 1 L of tap water
To find the minimum volume of potassium iodide needed to test a 500 mL sample of water, we multiply this value by the volume of the sample:
Minimum volume of potassium iodide = (0.00000705 L/L) * (500 mL / 1000 mL/L)
Minimum volume of potassium iodide = 0.00000353 L or 3.53 µL
Therefore, the minimum volume of a 0.00100 mol/L solution of potassium iodide that should be used to test a 500 mL sample of water is 3.53 µL.