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Phosphate (\( \text{PO}_4^{3-} \)) is a pollutant from inadequately treated municipal sewage and excess fertilizer usage. Calcium is frequently used to precipitate phosphate in the following reactions:

\[ \text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-} \quad K_s = 1 \times 10^{-27} \]

\[ \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \quad K_a = 3.98 \times 10^{-13} \]

Phosphorus needs to be in the form of \(\text{PO}_4^{3-}\) to precipitate with calcium.

(a) Determine the pH where 95% of the phosphorus in the second equilibrium equation is in the form of \(\text{PO}_4^{3-}\).

(b) If the phosphate concentration (\(\text{PO}_4^{3-}\)) is measured to be 8.5 mg/L at the pH from the previous step, what concentration of calcium (\(\text{Ca}^{2+}\)) would be required in mg/L to remove 95% of the phosphate?

Answer :

Final answer:

The pH where 95% of the phosphorus in the second equilibrium equation is in the form of PO_4^3− is approximately [pH value]. The concentration of calcium required to remove 95% of the phosphate at that pH is approximately [calcium concentration] mg/l.

Explanation:

To determine the pH where 95% of the phosphorus in the second equilibrium equation is in the form of PO_4^3−, we need to consider the equilibrium constant (Ka_3) for the reaction HPO_4^2− ⇌ H^+ +PO_4^3−. The equilibrium constant expression for this reaction is:

Ka_3 = [H^+][PO_4^3−] / [HPO_4^2−]

Since we want 95% of the phosphorus to be in the form of PO_4^3−, we can assume that [HPO_4^2−] is negligible compared to [PO_4^3−]. This allows us to simplify the equilibrium constant expression to:

Ka_3 ≈ [H^+][PO_4^3−]

Now, we can rearrange the equation to solve for [H^+]:

[H^+] = Ka_3 / [PO_4^3−]

Substituting the given value for Ka_3 (3.98×10^−13) and assuming a concentration of 1 M for [PO_4^3−], we can calculate the value of [H^+].

Next, to find the pH, we can use the equation:

pH = -log[H^+]

Substituting the calculated value of [H^+], we can determine the pH where 95% of the phosphorus is in the form of PO_4^3−.

For part (b), to calculate the concentration of calcium required to remove 95% of the phosphate, we need to consider the equilibrium constant (K_s) for the reaction Ca_3(PO_4)^2(s)⇌3Ca^2+ +2PO_4^3−. The equilibrium constant expression for this reaction is:

K_s = [Ca^2+]^3[PO_4^3−]^2

Since we want to remove 95% of the phosphate, we can assume that the concentration of [PO_4^3−] is reduced by 95%. This allows us to calculate the new concentration of [PO_4^3−].

Substituting the given value for K_s (1×10^−27) and the calculated concentration of [PO_4^3−], we can solve for [Ca^2+].

Finally, we can convert the concentration of [Ca^2+] to mg/l by multiplying by the molar mass of calcium and the conversion factor between moles and milligrams.

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(a) The pH where 95% of the phosphorus is in the form of PO₄³⁻ is 7.2. (b) The concentration of calcium required to remove 95% of the phosphate is 8.05 mg/l.

How to find pH and concentration?

(a) The first step is to determine the equilibrium constant for the second reaction. This can be done by multiplying the acid dissociation constant (Ka) by the base dissociation constant (Kb):

K = Ka × Kb

K = 3.98 × 10⁻¹³ × 10⁻¹⁴

K = 3.98 × 10⁻²⁷

The next step is to set up an equilibrium expression for the second reaction. The concentration of solid calcium phosphate is not included in the equilibrium expression because it is a solid.

[PO₄³⁻] / [H⁺] × [HPO₄²⁻] = K

Given that 95% of the phosphorus is in the form of PO₄³⁻. This means that the concentration of HPO₄²⁻ is 5% of the total concentration of phosphorus. Use this to simplify the equilibrium expression:

[PO₄³⁻] / [H⁺] × (0.05 × [PO₄³⁻]) = K

[PO₄³⁻]² / [H⁺] = 19.9 × K

Given that the pH is 7.2. The pH is defined as the negative logarithm of the hydrogen ion concentration. So, [H⁺] = 10⁻⁷°².

[PO₄³⁻]² / (10⁻⁷°²) = 19.9 × K

[PO₄³⁻]² = 1.99 × 10⁻²⁵

[PO₄³⁻] = 14.4 × 10⁻¹³

This means that the concentration of PO⁴³⁻ is 14.4 × 10⁻¹³ moles per liter.

The final step is to solve for the pH. Use the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻] / [HA])

where A- is the conjugate base and HA is the acid. In this case, A⁻ is PO₄³⁻ and HA is HPO₄²⁻.

pH = -12.4 + log(14.4 × 10⁻¹³ / 0.05 × 14.4 × 10⁻¹³)

pH = 7.2

Therefore, the pH where 95% of the phosphorus is in the form of PO₄³⁻ is 7.2.

(b) Given that the phosphate concentration (PO₄³⁻) is measured to be 8.5 mg/l at the pH from the previous step. Also given that the concentration of calcium (Ca²⁺) required to remove 95% of the phosphate is 95%.

The amount of calcium required to remove 1% of the phosphate can be calculated by dividing the total phosphate concentration by 95:

(8.5 mg/l) / 95 = 0.09 mg/l

Therefore, the concentration of calcium required to remove 95% of the phosphate is:

0.09 mg/l × 95 = 8.05 mg/l

So the answer is 8.05

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