Question

You want to prepare 450.0 mL of a 145 ppm K+ ion solution. The density of the solution is 1.01 g/mL. You have at your disposal a solution that is 0.728 M in K2SO4. How many mL of the K2SO4 solution do you need to use?

Answer 1

By definition, 1ppm = 1 mg solute per 1 kg solution

So, 145 ppm K^+ = 145 mg K^+ per 1 Kg solution.

Given that density = 1.01 g/mL

But, we know that 1 g/mL = 1 Kg/L

Thus, 1.01 g/mL = 1.01 kg/L

1 kg of solution= (1 kg) / (1.01 kg/L) = 0.9901 L

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Now, let us find moles of K^+ from the given mass of K^+.

Moles = Mass in g / Molar mass

145 mg K^+ = (0.145 g K^+) / (39.098 g/mol)

= 0.003709 moles of K^+
Now, let us calculate the molarity using this moles.

Molarity = Moles / Volume of solution

M of K^+ = (0.003709 moles of K^+) / (0.9901 L) = 0.003746 M

Next, we calculate the molarity of K^+ in the given stock solution:

Every mole of K2SO4 contains 2 mol of K^+ ions

(0.728 M in K2SO4) x (2 moles K^+/1 mole K2SO4) = 1.456 M

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Now, let us find the volume of K2SO4 required using the equation, M1V1 = M2V2

M1 = 1.456 M
V1 = ?
M2 = 0.003746 M
V2 = 450.0 mL

V1 = ((0.003746 M) x (450.0 mL)) / (1.456 M)
V1 = 1.158 mL