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?
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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:)
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