Question

If 200. mL of water is added to 300. mL of an aqueous solution that is 0.64 M in potassium sulfate, what is the concentration of potassium ions in the final solution?

Answer 1

Let us find out concentration of K2SO4 solution when water is
added to it ,using dilution formula

M1V1 =M2V2

Known numbers are

initial molarity M1 = 0.64 M
inital volume V1 = 300 ml
final molarity M2 = ??
final volume V2 = 200 ml + 300 ml = 500 ml

Thus final molarity M2 = M1V1/V2

Plug in all known numbers to get unknown concwntration[molarity]

M2 = 0.64 M*300ml/500 ml = 0.384 M

thus molarity of K2SO4 solution = 0.384 M

This means 1 liter solution has 0.384 moles of K2SO4

we have 1 mole K2SO4 dissociate in to 2 moles K+ ions in water;j;;the equation given below

[ K2SO4 -----> 2K+(aq) + SO42-(aq) ..is the dissociation reaction
in water]

Thus 0.384 moles K2SO4 has 2*0.384moles K+ ions

= 0.768 moles

So concentration of K+ ions = 0.768 moles/L

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:)