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October 26, 2014

October 26, 2014

Posted by **Shreya** on Friday, January 25, 2013 at 12:40am.

A. 2.2*10^-13

B. 6.2*10^-8

C. 1.6*10^-7

D. 4.5*10^-2

I dont get how to solve this.

- Chemistry 12 -
**Devron**, Friday, January 25, 2013 at 12:58amIf you are asking what is the dissociation constant (Kb) for HPO₄², you want be able to solve for it with the information given; you have to look up the Ka for HPO₄² in a book.

Ka for HPO₄²=4.2 x10^-13

Kw= Ka*Kb

Solving for Kb,

Kb=Kw/Ka= 1 x 10^-14/4.2 x10^-13 =2.40 x 10^-2

This answer is not one of your answer choices, so I'm a little stuck on this one.

- Chemistry 12 -
**DrBob222**, Friday, January 25, 2013 at 1:00amIt's Kw/k2 for H3PO4.

I'm sure you remember Kw = 1E-14.

- Chemistry 12 -
**Shreya**, Friday, January 25, 2013 at 1:03amHow do you know which ka to use? There is one on the base side of the chart, and one of the acid side of the chart. For the one of the acid side of the chart, the Ka is 2.2*10^-13 And for the one on the base side of the chart, it is 6.2*10^-8.

- Chemistry 12 -
**Devron**, Friday, January 25, 2013 at 1:04amWith that being said,

Kb=Kw/Ka= 1 x 10^-14/6.2 x10^-8 =1.6 x10^-7

- Chemistry 12 -
**Shreya**, Friday, January 25, 2013 at 1:06amThanks so much :) So if we were to find the Ka, then we would use the ka on the acid side of the chart?

- Chemistry 12 -
**DrBob222**, Friday, January 25, 2013 at 1:12amCorrect me if I'm wrong but

HPO4^2- + HOH ==> H2PO4^- + OH^-

k3 for H3PO4= 4.2E-13 = (H^+)(PO4^3-)/(HPO4^-) and that doesn't contain the ions needed for HPO4^2- to act as a base. k2 for H3PO4 does.

k2 = (H^+)(HPO4^2-)/(H2PO4^-)

I think the answer is c.

- Chemistry 12 -
**Devron**, Friday, January 25, 2013 at 1:13amYes, the way your reference is set-up. Remember, Kw= Ka*Kb. if you want to know the Kb, you need to know the Kw and Ka. If you want to know the Kb, you need to know the Ka and Kw.

- Chemistry 12 -
**Devron**, Friday, January 25, 2013 at 1:14amIts c, I used the wrong Ka value when I initially did it.

- Chemistry 12 -
**Shreya**, Friday, January 25, 2013 at 1:16amThanks soo much to both of you :)

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