Respuesta :
Answer:
[tex]=3.75M[/tex]
Explanation:
Hello,
In this case, given the concentration of the potassium phosphate, we can compute the concentration of potassium ions by noticing that in one mole of salt, three moles of potassium ions are present. Moreover, since the molar units (mol/L) are in terms of potassium phosphate we should apply the following mole-mole relationship:
[tex]1.25\frac{molK_3PO_4}{L}*\frac{3molK^+}{1molK_3PO_4} \\\\3.75\frac{molK^+}{L}=3.75M[/tex]
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Answer:
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.
