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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: banana1049 on May 13, 2007, 02:57:59 PM

Title: Solubility-product constant
Post by: banana1049 on May 13, 2007, 02:57:59 PM
What is the solubility-product constant of barium carbonate if a saturated solution is 1.1*10-4?
I know that Ksp=[Ba][CO2-3], but I'm not sure where to start with this.  Should I begin by balancing the equation for BaCO3 and then find the concentration of each part?

Just so you know, I'm not asking anyone to give me the answer to this question.  I have a final next week and I need to know how to understand problems like this.  I'm just asking for a little clarification because we raced through the last three chapters in class and I didn't quite understant everything.
Title: Re: Solubility-product constant
Post by: Borek on May 13, 2007, 03:27:10 PM
If saturated solution is 1.1*10-4M, what are concentrations of Ba2+ and CO32-?
Title: Re: Solubility-product constant
Post by: banana1049 on May 13, 2007, 04:12:32 PM
Ok I ended up with

.0022g BaCO3/100g H2O * 1g H20/1ml * 1000ml/1L *1molBaCO3/197.34g  = 1.1*10-4

then [Ba] = [CO3] = 1.1*10-4
Ksp = (1.1*10-4)2 =
1.2*10-6
Title: Re: Solubility-product constant
Post by: Borek on May 13, 2007, 04:57:02 PM
Quote from: banana1049 on May 13, 2007, 04:12:32 PM
.0022g BaCO3/100g H2O * 1g H20/1ml * 1000ml/1L *1molBaCO3/197.34g  = 1.1*10-4

No idea why and what, but looks OK.

Quote
then [Ba] = [CO3] = 1.1*10-4

OK

Quote
Ksp = (1.1*10-4)2 = 1.2*10-6

Either error or typo.
Title: Re: Solubility-product constant
Post by: banana1049 on May 13, 2007, 10:19:21 PM
Quote
Ksp = (1.1*10-4)2 = 1.2*10-6

Either error or typo.

[/quote]

My calculator keeps giving me 1.21*10-6 but that's not right.  I think it's actually 1.2*10-8.
Title: Re: Solubility-product constant
Post by: english on May 13, 2007, 11:22:32 PM
Well your mistake is simply algebraic.  You're right.