What is the concentration of potassium ions in 153 mLof a 1.25 M K3PO4 solution

The concentration of potassium (K⁺) ions in the solution of 153 mL of 1.25 M K₃PO₄ is 3.75 moles

Explanation:

The parameters given are;

Volume of K₃PO₄ = 153 mL = 0.153 L

Concentration of K₃PO₄ = 1.25 M

The ionization of K₃PO₄ is as follows;

K₃PO₄ ⇄ 3·K⁺ (aq) + PO₄³⁻ (aq)

Therefore, 1 mole of K₃PO₄, produces 3 moles of potassium (K⁺) ions

Hence, the number of moles of K₃PO₄ in 153 mL of 1.25 M K₃PO₄ is found as follows;

153 mL of 1.25 M K₃PO₄ = 0.153 L of 1.25 M K₃PO₄

The number of moles present in 0.153 L of 1.25 M K₃PO₄ = 0.153 × 1.25 moles

The number of moles present in 0.153 L of 1.25 M K₃PO₄ = 0.19125 moles of K₃PO₄

Since 1 mole of K₃PO₄ produces 3 moles of potassium (K⁺) ions we have;

0.19125 moles of K₃PO₄ will produce, 3 × 0.19125 moles of potassium (K⁺) ions which gives;

0.19125 moles of K₃PO₄ will produce 0.57375 moles of potassium (K⁺) ions

Therefore, the amount of moles of potassium (K⁺) ions in 153 mL = 0.57375 moles

The the amount of moles of potassium (K⁺) ions in 1 L (1000 mL) = (1000/153) × 0.57375 moles

The the amount of moles of potassium (K⁺) ions in 1 L = 3.75 moles

Therefore, the concentration of potassium (K⁺) ions in the solution of 153 mL of 1.25 M K₃PO₄ = 3.75 moles.

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