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 It is no longer valid to approximate that at equilibrium is the same as its initial value.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

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.