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–.

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 =

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