Question

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−?

Answer 1

The acid ionization constant (Ka) for the HSO4− ion is calculated using the equilibrium concentrations provided to be 1.2 × 10−2. This constant is determined from the equation Ka = [H3O+][SO42−] / [HSO4−].

Step-by-step explanation:

The acid ionization constant (Ka) for a weak acid can be determined using the equilibrium concentrations of its ions and molecules. In the case of the HSO4− ion, which is a weak acid, the equation for its dissociation into hydronium ions (H3O+) and sulfate ions (SO42−) is:

HSO4−(aq) ⇌ H3O+(aq) + SO42−(aq)

The Ka expression for this reaction is given by:

Ka = [H3O+][SO42−] / [HSO4−]

Using the provided equilibrium concentrations:

Ka = (0.027 M)(0.13 M) / 0.29 M

This simplifies to:

Ka = 0.00351 / 0.29

Ka = 1.2 × 10−2

This is the acid ionization constant for the HSO4− ion at the given equilibrium concentrations.

Answer 2

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.

Step-by-step 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.
`\text{K}_a = 1.3* 10^(-2)` 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⁻.

`\begin{array}ccccc\text{R}&{\text{HSO}_4}^(-) &\rightleftharpoons& \text{H}^(+)&+&{\text{SO}_4}^(2-)\\\text{I}&0.14&&&&\\\text{C}&-x & &+x &&+x\\\text{E}&0.14 - x&&x&&x\end{array}`.

`\text{K}_a = 1.3 * 10^(-2)` for
`{\text{HSO}_4}^(-)`. As a result,

`\displaystyle \frac{[\text{H}^(+)]\cdot[{\text{SO}_4}^(2-)]}{[{\text{HSO}_4}^(-)]} = \text{K}_a`.

`\text{K}_a` is large. It is no longer valid to approximate that
`[{\text{HSO}_4}^(-)]` at equilibrium is the same as its initial value.

`\displaystyle (x^(2))/(0.14 - x) = 1.3* 10^(-2)`.

`x^(2) + 1.3* 10^(-2)\;x - 0.14* 1.3* 10^(-2) = 0`.

Solve the quadratic equation for
`x`.
`x\ge 0` since
`x` represents a concentration.

`x = 0.0366538`.

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

• 
  `[{\text{SO}_4}^(2-)] = x = 0.037\;\text{mol}\cdot\text{L}^(-1)`;
• 
  `[{\text{H}^(+)] = x = 0.037\;\text{mol}\cdot\text{L}^(-1)`;
• 
  `[{\text{HSO}_(4)^(-)] = 0.14 - x = 0.10\;\text{mol}\cdot\text{L}^(-1)`.