Question

4. Describe (in detail) using both words and calculations how you would prepare 2.0 L of a 0.100 M KCl solution from a 5.5 M stock KCl solution. (2 pt) 5. Aluminum hydroxide will react with sulfuric acid to yield aluminum sulfate and water according to the equation below: (4 pt) 3H2SO4(aq) + 2Al(OH)3(aq) → Al(SO4)3(aq) + 6H20(1) a) How many grams of aluminum hydroxide will react with 75,0 mL of 0.319 M sulfuric acid solution? b) If 25.26 mL of a 0.125 M solution of aluminum hydroxide reacts exactly with 22.00 mL of a sulfuric acid solution, what is the molarity of the sulfuric acid solution?

Answer 1

4)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 5.5 M

M2 = 0.1 M

V2 = 2.0 L = 2000 mL

use:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (0.1*2000)/5.5

V1 = 36 mL

So, we take 36 mL of 5.5 M stock solution and add enough water to it until final volume is 2.0 L

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