Question

If 87 grams of K2SO4 (molar mass 174 grams) is dissolved in enough water to make 250 milliliters of solution, what are the concentrations of the potassium and the sulfate ions?

Answer 1

Answer:

$M_{K^+}=4.0M \\\\M_{SO_4^{2-}}=2.0M$

Explanation:

Hello!

In this case, since the molarity of a solution is defined as the moles of solute divided by the volume of solution liters, given the mass of potassium sulfate, we can compute the moles by using its molar mass (174.24 g/mol):

$n_{K_2SO_4}=87gK_2SO_4*\frac{1molK_2SO_4}{174.27gK_2SO_4} =0.50molK_2SO_4$

Thus, since one mole of potassium sulfate has two moles of potassium ions (K₂) and one mole of sulfate ions, we can compute the moles of each ion as shown below:

$n_{K^+}=0.50molK_2SO_4*\frac{2molK^+}{1molK_2SO_4}=1.0molK^+\\\\ n_{SO_4^{2-}}=0.50molK_2SO_4*\frac{1molSO_4^{2-}}{1molK_2SO_4}=0.50molSO_4^{2-}$

In such a way, the molarity of each ion turns out:

$M_{K^+}=\frac{1.0mol}{0.250L}=4.0M \\\\M_{SO_4^{2-}}=\frac{0.5mol}{0.250L}=2.0M$

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