Powdered AgOH is slowly added to 2.0 L of a 0.020 M solution of NaOH until no more

AgOH will dissolve. If Ksp = 1.4 x 10 -8 for AgOH:
a. What is [Ag
+] in the saturated solution?
b. What mass of AgOH had to be added to the solution to produce this
concentration of Ag+?

Ksp = (Ag^+)(OH^-)

You know Ks and OH^-, substitute and solve for (Ag^+) (in mols/L).

b. So you know (Ag^+) in mols/L. In 2.0L you will have twice as many mols. Then grams = mols x molar mass = ?

thanks

but how is the concentration of OH determined?

To solve this problem, we need to understand the concept of solubility product constant (Ksp) and how it relates to the solubility of a compound in water.

a. What is [Ag+] in the saturated solution?

Since AgOH is considered fully dissociated in water, the solubility of AgOH can be represented as the concentration of Ag+ ions. Let's assume that x mol/L of Ag+ ions are formed in the saturated solution.

The balanced equation for the dissociation of AgOH is:

AgOH ⇌ Ag+ + OH-

We can write the solubility product expression using the Ksp value:

Ksp = [Ag+][OH-]

Since AgOH is a strong base, the concentration of OH- ions in the solution will be equal to the concentration of NaOH initially present.

Given that the initial concentration of NaOH is 0.020 M, the concentration of OH- ions is also 0.020 M.

Now, we can substitute the known values into the Ksp expression:

1.4 x 10^-8 = x * 0.020

Solving for x gives:

x = (1.4 x 10^-8) / 0.020

x ≈ 7 x 10^-7 M

Therefore, the concentration of Ag+ ions in the saturated solution is approximately 7 x 10^-7 M.

b. What mass of AgOH had to be added to the solution to produce this concentration of Ag+?

To calculate the mass of AgOH, we need to use the molar mass of AgOH, which is 123.87 g/mol.

Since the compound is fully dissociated, the number of moles of AgOH will be equal to the number of moles of Ag+ ions in the solution.

To find the number of moles of Ag+, we multiply its molar concentration by the volume of the solution:

moles of Ag+ = concentration of Ag+ * volume of the solution

moles of Ag+ = (7 x 10^-7 M) * (2.0 L)

moles of Ag+ = 1.4 x 10^-6 mol

Now, we can convert the moles of Ag+ ions to grams of AgOH using its molar mass:

mass of AgOH = moles of Ag+ * molar mass of AgOH

mass of AgOH = (1.4 x 10^-6 mol) * (123.87 g/mol)

mass of AgOH ≈ 1.7 x 10^-4 g

Therefore, approximately 1.7 x 10^-4 grams of AgOH had to be added to the solution to produce a concentration of Ag+ ions of approximately 7 x 10^-7 M.