Making A Solution

Question

Make a 50.0 ml of a solution that is 0.080 M sulfuric acid and also 0.20 M in KIO3.

Calculate how many milliliters of 0.080 M sulfuric acid are needed to make 50.0 ml of 0.080 M sulfuric acid. Also, calculate how many grams of potassium iodate are needed to make 50.ml of 0.20 M in KIO3.

Answer 1

Hello Brook

a) Make a 50.0 ml of a solution that is 0.080 M sulfuric acid and also 0.20 M in KIO3.

Calculate moles of H2SO4 and KIO3 using the volume and molarity. Then use their respective molar masses to calculate how many g are needed to prepare the solution.

mol H2SO4 = 0.080 mol/L x 50.0 mL x (1 L / 1000 mL) = 0.00400 mol

g H2SO4 = 0.00400 mol x (98 g H2SO4 / 1 mol H2SO4) = 0.392 g H2SO4

mol KIO3 = 0.20 mol/L x 50.0 mL x (1 L / 1000 mL) = 0.0100 mol

g KIO3 = 0.0100 mol x (214 g KIO3 / 1 mol KIO3) = 2.14 g KIO3

b) This is a dilution question where the concentration of the "end" solution decreases. However, the starting and end concentration of sulfuric acid listed are the same (0.080 M), thus, there is an error with one of the numbers in your question.

c) See above KIO3 calculation for this answer.