Question: When 1.97 grams of Pb(OH)2(s) is added to 1.00 liter of pure water at 25 C a saturated solution (some solid is present) is formed that has a pH of 8.97. Compute the Ksp value for Pb(OH)2 using the data.

[OH]= 9.334e-6 and the [Pb^+2]= 9.334e-6/2= 4.667e-6. So the ksp= 9.334e-6 * 4.667e-6= 4.356e-11

You didn't square the OH^-

Right! so after i square it the correct answer would be 4.066e-16

I think so. Look up Ksp in your set of Ksp tables and see how close that comes. I have an old quant book (about 15 years old) that shows 2.5 x 10^-16.

To compute the Ksp value for Pb(OH)2 using the given data, you need to know the concentration of hydroxide ions ([OH-]) and the concentration of lead ions ([Pb^+2]).

We are given the [OH-] as 9.334e-6. This represents the concentration of hydroxide ions in the saturated solution.

To find the concentration of lead ions, we divide the [OH-] by 2, as the balanced chemical equation for the dissociation of Pb(OH)2 is:

Pb(OH)2(s) ⇌ Pb^+2(aq) + 2OH-(aq)

Therefore, the [Pb^+2] is calculated as 9.334e-6 / 2 = 4.667e-6.

Finally, we can calculate the Ksp value using the equation:

Ksp = [Pb^+2] * [OH-]

Substituting the values, we get:

Ksp = 4.667e-6 * 9.334e-6

Calculating this expression yields the Ksp value of Pb(OH)2 as 4.356e-11.