Consider the dissolution equation of lead(II) chloride.
PbCl2 (s) ⇆ Pb2+ (aq) + 2Cl− (aq)
Suppose you add 0.2307 g of PbCl2 (s) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of Pb2+ (aq) is 0.0159 M and the concentration of Cl− (aq) is 0.0318 M.
What is the value of the equilibrium constant, Ksp, for the dissolution of PbCl2?

The solubility product of lead(II) chloride is .

Explanation:

Concentration of lead (II) ions =

Concentration of chloride ion =

The expression of a solubility product will be given as:

The solubility product of lead(II) chloride is .

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.