# what is the concentration of k in 0.15 m of k2s?

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## what is the concentration of k in 0.15 m of k2s?

#### 1) Answer is: the concentration of potassium cations (K⁺) is 0.3 M.

Balanced chemical reaction (dissociation) of potassium sulfide in the water:

K₂S(aq) → 2K⁺(aq) + S²⁻(aq).

c(K₂S) = 0.15 M.

From balanced reaction: n(K₂S) : n(K⁺) = 1 : 2, because volume of the solution is constant: c(K⁺) = 2 · c(K₂S).

c(K⁺) = 0.3 M; concentration of potassium cations.

#### 2) Answer is: its concentration is half that of the Cl⁻ ion.

Balanced chemical reaction (dissociation) of calcium chloride (CaCl₂) in the water:

CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq).

From balanced reaction: n(Ca²⁺) : n(Cl⁻) = 1 : 2, because volume of the solution is constant: c(Ca²⁺) = c(Cl⁻) ÷ 2.

n is amount of the substance.

#### 3) Answer is: 46.176 grams of Na₃PO₄ will be needed.

Balanced chemical reaction (dissociation) of sodium phosphate  (Na₃PO₄) in the water:

Na₃PO₄(aq) → 3Na⁺(aq) + PO₄³⁻(aq).

V = 650 mL ÷ 1000 mL.

V = 0.650 L; volume of the solution.

c(Na⁺) = 1.3 M; concentratium of sodium cations. what is the concentration of k in 0.15 m of k2s?

From balanced reaction: n(Na₃PO₄) : n(Na⁺) = 1 : 3, because volume of the solution is constant: c(Na₃PO₄) = c(Na⁺) ÷ 3.

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c(Na₃PO₄) = 1.3 M ÷ 3.

c(Na₃PO₄) = 0.43 M.

n(Na₃PO₄) = c(Na₃PO₄) · V(solution).

n(Na₃PO₄) = 0.43 M · 0.65 L.

and

n(Na₃PO₄) = 0.281 mol.

m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).

m(Na₃PO₄) = 0.281 mol · 163.94 g/mol.

then

m(Na₃PO₄) = 46.176 g. what is the concentration of k in 0.15 m of k2s?

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