Tuesday

September 23, 2014

September 23, 2014

Posted by **LT** on Monday, April 14, 2008 at 8:06pm.

The Solubility Product Constant PbSO4

Ksp= 1.8 x 10^-8

The Ksp expression is:

Na2SO4 = [2Na][SO4^-2]

[2Na]=2x

[SO4^-2]=x

Ksp= 1.8 x 10^-8 = (x)(2x)

Ksp= 1.8 x 10^-8 /4x^3

1.8 x 10^-8/4 = x^3

4.5 x 10^-9 = x^3

is this right so far... what do i do next please

- Chemistry -
**DrBob222**, Monday, April 14, 2008 at 8:11pmNo, you have made some wrong assumptions. First, PbSO4 is the insoluble material. Na2SO4 is soluble in water. Second, this is a common ion problem in which the presence of the sulfate ion from Na2SO4 will decrease the solubility of PbSO4.

- Chemistry -
**LT**, Monday, April 14, 2008 at 8:25pmSo what do i do then to solve for the ksp

- Chemistry -
**DrBob222**, Monday, April 14, 2008 at 8:33pmThe problem doesn't ask for you to solve for Ksp. It asks for the solubility of PbSO4 in 1 x 10^-3 M Na2SO4.

PbSO4 ==> Pb^+2 + SO4^=

Ksp = (Pb^+2)(SO4^=) = 1.8 x 10^-8

If solubility of PbSO4 = S (or x if you wish), then (Pb^+) = S and (SO4^=) = S

The Na2SO4 in solution is

Na2SO4 ==> 2Na^+ + SO4^=

Since it is 1 x 10^-3 M in Na2SO4, then (SO4^=) = 1 x 10^-3.

Now set those up in the Ksp expression and solve for S. Remember that the total sulfate is that from the PbSO4 + that form the Na2SO4.

- Chemistry -
**LT**, Monday, April 14, 2008 at 8:35pmSo im solving for two expressions... is that it.

- Chemistry -
**LT**, Monday, April 14, 2008 at 8:48pmok I got x^2 = 1.8 x 10^-8 = 1.34 x 10^-4 for PbSO4 is this right

- Chemistry -
**DrBob222**, Monday, April 14, 2008 at 8:51pmYou are solving for the only unknown in the equation, which is S.

Ksp = (Pb^+2)(SO4^-2) = 1.8 x 10^-8

(S)(S + 1E-3)= 1.8 x 10^-8

S =

- Chemistry -
**LT**, Monday, April 14, 2008 at 8:56pmi subtracted 1E-3 from the 1.8E-8 and it got me a -9.999E-4 then i took the sqrt cause (X)(X) is x^2 and then the calculator told me an error

- Chemistry -
**DrBob222**, Monday, April 14, 2008 at 9:54pmKsp = (Pb^+2)(SO4^=) = 1,8 x 10^-8

(Pb^+2) = x

(SO4^=) = x+0.001

(x)(x+0.001) = 1.8 x 10^-8

You can see that this is a quadratic equation and we could solve it as a quadratic. The exact solution is (and I'm switching to y so as not to confuse the x with the times sign),

y^2 + 0.001y = 1.8 x 10^-8

There are two ways to do this.

1. Simplify the equation so there is no quadratic. You do that by making the assumption that y + 0.001 = 0.001; that is, that y is such a small number that a small number + 0.001 is still 0.001. If we do that then we have

y = 1.8 x 10^-8/0.001 = 1.8 x 10^-5 M. We need to test to see if our assumption is ok. 0.001 + 1.8 x 10^-5 = 0.00102 and there is just a small difference so we let it go and work it the easy way.

2. we can solve the quadratic by equating

y^2 + 0.001 -1.8 x 10^-8 = 0

I solved this by the quadratic formula and obtained 1.77 x 10^-5 so you see we don't make a bad assumption by solving it the easy way.

**Answer this Question**

**Related Questions**

Chemistry! URGENT! - Calculate the solubility, in molL―1, of lead sulfate...

chemistry - what is the solubility of barium sulfate in a solution containing 0...

chemistry - Calulate the molar solubility of Al(OH)3 in water. (Ksp = 3 x 10^-34...

Chemistry(Please check) - What is molar solubility of Al(OH)3 at 25 degrees C ...

chemistry - The solubility product constant (Ksp) for the dissolution of PbSO4 ...

science - In an experiment to determine the solubility of lead chloride (PbCl2...

chem - The solubility product of silver sulfate is 1.6* 10^-5. What is the molar...

Chemistry - Tooth enamel is made up of hydroxyapatite, Ca_10(PO_4)_6(OH)_2 or ...

chemistry - Solid Na2SO4 is added to a solution which is 0.014 M in Pb(NO3)2 and...

chem - The molar solubility of MgF2 at 25°C is 1.2 10-3 mol/L. Calculate Ksp. i...