What are the concentrations of hso4−, so42−, and h+ in a 0.14 m khso4 solution? (hint: h2so4 is a strong acid; ka for hso4− = 1.3 10-2.) hso4−?

At equilibrium:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

All three values are in 2 sig. fig. as in Ka or the KHSO₄ concentration.

Explanation:

There's initially no SO₄²⁻ and negligible amount of H⁺ in the solution. KHSO₄ is a salt. All KHSO₄ will dissociate to form K⁺ and HSO₄⁻ ions. The initial concentration of [HSO₄⁻] will be 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

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for . As a result,

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is large. It is no longer valid to approximate that at equilibrium is the same as its initial value.

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Solve the quadratic equation for . since represents a concentration.

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The least significant number in the question comes with 2 sig. fig. Round the results to 2 sig. fig:

acid rain does not usually kill trees directly. instead, it is more likely to weaken the trees by damaging their leaves, limiting the nutrients available to them, or poisoning them with toxic substances slowly released from the soil. the main atmospheric pollutants that affect trees are nitrates and sulphates.