A) A saturated solution of lead(II) fluoride, , was prepared by dissolving solid
in water. The concentration of
ion in the solution was found to be
. Calculate
for
.
for
is
.
D) What is the pH of a buffer prepared by adding 0.809 of the weak acid
to 0.305
of
in 2.00
of solution? The dissociation constant
of
is
. What is the
after 0.150
of
is added to the buffer from Part A? Assume no volume change on the addition of the acid. What is the
after 0.195
of
is added to the buffer from Part A? Assume no volume change on the addition of the base.
Expert Answer
[Pb2+] = 2.08 * 10^-3 M
One mole of Pb F2 gives two moles of fluoride ions
so [F-] = 2 * 2.08 * 10^-3 M
Ksp = ionic product of a saturated solution
= [Pb^2+] [ Cl -]^2
= (2.08 * 10^-3 M) * (2.08 * 10^-3 M)^2
= 9.00 * 10^-9
Solubility of silver sulfate can be calculated using the Ksp formula:
Ksp = [Ag2+] [SO4^2-]
Hope this helps