Question

Calculate the concentration of silver ions that would result if a saturated solution of silver sulfate, AgSO4 (Ksp = 1. 5e-5) was made. (ignore sulfate ion hydrolysis)

a) 2. 74e-3M

b) 3. 87e-3M

c) 1. 55e-2M

d) 3. 11e-2M

Answer 1

The concentration of silver ions in a saturated solution of silver sulfate, where the Ksp is 1.5e-5, is calculated as the square root of the Ksp. This gives a [Ag+] concentration of 3.87e-3M, which is answer (b).

Step-by-step explanation:

To calculate the concentration of silver ions, [Ag+], in a saturated solution of silver sulfate, AgSO4, we will use the solubility product constant (Ksp). We know that the dissolution of AgSO4 in water can be represented by the following equilibrium:

AgSO4(s) → Ag+(aq) + SO42-(aq)

The solubility product expression for this equilibrium is:

Ksp = [Ag+][SO42-]

Since AgSO4 dissociates to give one Ag+ ion and one SO42- ion per formula unit that dissolves, we can assume that the concentration of Ag+ and SO42- are equal when the solution is saturated. We let the concentration of Ag+ ions be 's', then the equation will be:

Ksp = s * s = s^2

Solving for 's' gives us the concentration of Ag+ in the solution:

s = sqrt(Ksp) = sqrt(1.5e-5) = 3.87e-3 M

Thus, the correct answer is (b) 3.87e-3M, which is the concentration of silver ions in a saturated solution of silver sulfate.

Answer 2

Answer: The concentration of silver ions will be 3.87e-3 M.

Step-by-step explanation:

Consider the balanced chemical equation for the dissolution of
`\text{AgSO}_4` in pure water:

`\text{AgSO}_4 \text{ (s)} \rightleftharpoons \text{Ag}^(2+) \text{ (aq)} + \text{SO}_4^(2-) \text{ (aq)}`

The value of Ksp for silver sulfate is given as
`1.5 * 10^(-5)`. We know that the expression for Ksp for the above equation is:

`K_(sp) = [\text{Ag}^(2+)][\text{SO}_4^(2-)]`

The concentration of the silver ions and sulfate ions are unknown, but they must be equal according to the chemical reaction. Then, replacing Ksp with the known value, and replacing the concentrations with "x" gives:

`1.5 * 10^(-5) = (x)(x) = x^2 \rightarrow 1.5 * 10^(-5) = x^2 \rightarrow x = \sqrt{1.5 * 10^(-5)} \approx 0.00387 \text{ M}`

Therefore, the concentration of silver ions will be:

`[\text{Ag}^(2+)] = x = 0.00387 \text{ M} = 3.87 * 10^(-3) \text{ M}`