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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sierrastudent on December 09, 2006, 08:18:11 PM

Title: principles of reactivity: other aspects of aqueous equilibria
Post by: sierrastudent on December 09, 2006, 08:18:11 PM

the ka's for H₃AsO₄ are: K1=2.5x10^-4         K2=5.6x10^-8        K3=3.0x10^-13
calculate the pH of the solution formed when 60.0 mL of 0.12 M KH2AsO4 is mixed with:
a. 40.0 mL of 0.15 M K₂HAsO₄.
I believe the net ionic equation is H₂AsO₄^- >H⁺ + HAsO₄^2-.
Is this correct? I believe I can do this problem if I get the net ionic equations, but I seem to be having a problem with this. Thank you for the net ionic equation :).  So I can get my bearings straight, what is the whole equation when these two solutions are added because I want to see where both sides cancel to get that net ionic equation? K⁺ +H₂AsO₄^- + 2K⁺ + HAsO₄^2- ----> ?

b. 140. mL of 0.060 M H₃AsO₄.

Title: Re: Principles of reactivit: other aspects of aqueous equilibria
Post by: Borek on December 09, 2006, 08:31:43 PM

Please read forum rules (http://www.chemicalforums.com/index.php?page=forumrules).

Title: Re: principles of reactivity: other aspects of aqueous equilibria
Post by: Yggdrasil on December 09, 2006, 10:44:39 PM

Yes, the correct net ionic equation to use for the acid/base equilibrium in (a) will be:

H₂AsO₄^- <--> HAsO₄^2- + H⁺

You might find it useful, to use the Henderson-Hasselbalch equation (http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation)