Chemistry - Ksp / Molarity Question Help

Ksp Chemistry Questions?

Ag3PO4 at equilibrium 3 Ag+ plus PO4^-3

from the balanced equation you can see that for each mole of silver phosphate that dissolves, three moles of Ag+ will go into solution along with one mole of PO4^-3 ion.

so we triple the given solubility for silver ion and leave the solubility for phosphate the same as given.

Then we raise each molarity to the power of its coefficient in the equation above.

Molarity Ag+ = 3 x l.8 x l0^-5 = 5.4 x l0^-5

Molarity P04^-3 = l.8 x l0^-5

Ksp = (5.4 x l0^-5)^3(l.8 x l0^-5) = 2.83 x l0^-18th

Ksp is the product of the molar solubility of each ion raised to the power of its coefficient in the equation.
Ksp values are used to express the solubility of slightly soluble substances. The higher the Ksp value, the more soluble the substance.

Ksp BaSO3 = (Ba++)(SO3=)

In this solution we know the molarity of Ba++ is .l0 molar

And we know that BaSO3 must not be very soluble in water by looking at its low solubilty above

and since BaSO3 is not very soluble, it must be at equilibrium like this

BaSO3(s) at equilib with Ba++(aq) plus SO3=(aq)

If only 8.0 x l0^-6 moles per liter of BaSO3 dissolved, this means that the same molarity of SO3= was created in the process of dissolving.

And of course the molarity of the Ba++ is an overwhelming .l0 molar

so Ksp = (.10 molar Ba++) ( 8.0 x l0^-6 molar SO3=) = 8.0 x l0^-7 The tiny molarity of Ba++ created in the dissolving process can be ignored.

Ksp and Molarity Question?

28.7 mL of 0.199 M Ca2+ solution is titrated to an equivalence point with a solution containing PO43–. It requires 12.6 mL of the phosphate solution to reach the equivalence point. What is the molarity of the phosphate solution?

The answer choices are:

0.453 M
0.907 M
0.151 M
0.302 M

An explanation would be appreciated, thanks so much!

Chem Question - Ksp/Molarity!?

The Ksp of Ca(OH)2 in water at 25°C is approximately 6.5 ✕ 10−6. Assuming perfect dissociation, calculate the molarity of the OH − ion in a saturated calcium hydroxide solution.

Thanks!

Chemistry - Ksp / Molarity Question! Please help!?

The Ksp of Ca(OH)2 in water at 25°C is approximately 6.5 ✕ 10−6. Assuming perfect dissociation, calculate the molarity of the OH − ion in a saturated calcium hydroxide solution.

Thanks!

Chemistry solubility question? Please help?

The main equilibrium is:
BaSO4 (s) <==> Ba++ (aq) + SO4-- (aq)
The addition of K2SO4 shifts the equilibrium to the left; that is, BaSO4 will become less soluble.
The equilibrium equation is:
Ksp = [Ba++] [SO4--] = 1.1x10^-10.
Let x moles of BaSO4 dissolve in a liter of water. Since there are 0.800 moles of SO4-- already dissolved in the liter of water, the equation becomes:
1.1x10^-10 = x * (x + 0.800).
If you wish, you can solve for x here by rearranging to the quadratic and using the quadratic formula, but there is an easier way. Since Ksp is such a small number, x will be too. So 0.800 >> x, and therefore the expression in the parentheses can be closely approximated by 0.800. The final equation becomes:
1.1x10^-10 = x * 0.800,
which solves easily to x = 1.4x10^-10 M.

Chemistry help? Calculating molarity given Ksp?

So for every one mole of dissociation of solid that occurs, 1 mole of Ca and 2 moles of OH- will be formed. A mathematical equation representing this according to solubility equilibria would be:
6.5 x 10^-6 =[Ca2+][OH-]^2

6.5 x 10^-6 =(x)(2x)^2

6.5 x 10^-6 =(x)(4x^2)

and solve for x!

Chemistry question about Ksp and Solubility Product?

Ce(IO3)3 at equilibrium with Ce+3 + 3IO3-

Initial concentration of Ce after adding the two solutions together = .0125 Molar
( Adding the two solutions together cuts each concentration in half)

and initial concentration of IO3- wll be .02 Molar

Trial Ksp with these concentrations. Ksp = (Ce+3)(IO3-)^3

Trial Ksp = (.0125)(.02)^3 = 1 x l0^-7 far larger than the actuall Ksp so a precipitate will occur.

Now lets compare molar concentrations of ions with the balanced equations reacting ratio
.0125M >02 M

Ce(IO3)3(s) at equilibrium Ce+3(aq) + 3 IO3-

Since it takes 3 moles of IO3- for each mole of Ce+3 to form a precipitate, by inspection, we will react essentially all of the IO3- ions in soution.

In other words, .02 Molarl IO3- will react with only one third the molarity of Ce+3 ions.

so .02 over 3 = .0067 moles per liter Ce+3 reacted and formed into additional precipitate.

so remaining Ce+3 molarity should be .0125 moles per liter Ce+3 minus .0067 moles per liter lost

= .006 Moles per liter remaining Ce+3 in solution.

*URGENT* Ksp Chemistry problem!! HELP!?

moles NaOH used = M NaOH x L NaOH = (0.1013)(0.01586) = 0.001607 moles NaOH

NaOH + KHP ==> H2O + NaKP

Since KHP and NaOH react in a 1:1 mole ratio, then moles NaOH = moles KHP.

molarity KHP solution = moles KHP / L of KHP solution = 0.001607 / 0.00300 = 0.536 M

The dissolution reaction for KHP: KHP(s) ==> H+(aq) + KP-(aq)

So 0.536 M KHP would produce 0.536 M H+ and 0.536 M KP-.

Ksp = [H+][KP-] = (0.536)(0.536) = 0.287

The temperature of the titration wouldn't matter too much. If the temperature was higher during the titration, the KHP that already dissolved will stay in solution. If the titration temperature were cooler, some KHP could precipitate from solution but the solid would still react with the NaOH.

How do I calculate Ksp when given the Molarity of a solution?

MgF2 dissolves according to the equation:

MgF2(s) <--> Mg2+(aq) + 2 F-(aq)

So, Ksp will have the form:

Ksp = [Mg2+][F-]^2.

In order to calculate the value of Ksp, just plug in the concentrations and calculate it.