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1) Determine the grams of phosphate in 0.430 g potassium dihydrogen phosphate and show that the resulting solution when 0.430 g potassium dihydrogen phosphate is dissolved in 100 ml (100 g) water is 3000 ppm (3000 mg/L) in PO43–.

Respuesta :

jufsanabriasa

Answer:

3000ppm of PO₄³⁻

Explanation:

First, we need to calculate the moles of potassium dihydrogen phosphate, KH₂PO₄. These moles are equal to moles of phosphate, PO₄³⁻:

Moles KH₂PO₄ -Molar mass: 136.086g/mol-:

0.430g KH₂PO₄ * (1mol / 136.086g) = 3.16x10⁻³moles KH₂PO₄ = Moles PO₄³⁻.

Now, mass of PO₄³⁻ is:

Mass PO₄³⁻ -molar mass: 94.97g/mol:

3.16x10⁻³moles PO₄³⁻ * (94.97g / 1mol) = 0.300g of PO₄³⁻ = 300mg

In 100mL = 0.100L

300mg PO₄³⁻ / 0.100L =

3000ppm of PO₄³⁻

dsdrajlin

0.430 g of KH₂PO₄ contain 0.300 g of PO₄³⁻. When dissolved in 100 mL of water, the resulting solution is 3000 ppm in PO₄³⁻.

We want to calculate the mass of PO₄³⁻ in 0.430 g of  KH₂PO₄.

We know that:

  • The molar mass of PO₄³⁻ is 94.97 g/mol.
  • The molar mass of KH₂PO₄ is 136.08 g/mol.
  • The mass ratio of PO₄³⁻ to KH₂PO₄ is 94.97:136.08.

0.430 g KH₂PO₄ × (94.97 g PO₄³⁻/136.08 g KH₂PO₄) = 0.300 g PO₄³⁻

When 0.300 g  (300. mg) of PO₄³⁻ are dissolved in 100 mL of water, the resulting concentration, in ppm, is:

300. mg/0.100 L = 3000 ppm

0.430 g of KH₂PO₄ contain 0.300 g of PO₄³⁻. When dissolved in 100 mL of water, the resulting solution is 3000 ppm in PO₄³⁻.

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