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April 24, 2014

April 24, 2014

Posted by **kelly** on Wednesday, October 26, 2011 at 7:36pm.

- chemistry -
**DrBob222**, Wednesday, October 26, 2011 at 8:41pmpH = pKa + log[(base)/(acid)]

Convert Kb for NH3 into Ka and pKa, substitute into the H-H equation (1.70 for (base), and solve for (acid) in moles/L. Then convert to grams NH4Cl in 0.490 L. Post your work if you get stuck.

- chemistry -
**kelly**, Wednesday, October 26, 2011 at 9:15pmI still have it wrong. I have pKa = 9.25. Solved for acid and got 2.39. Then I converted to grams and 62.25. Here's my work ka = 1.0e-14/1.8e-5 = 5.56e-10. solved for pKa -log(5.56e-10) = 9.25. H-H = 9.40=9.25 + log(1.70/acid). acid = 2.39. then, 2.39 x 0.490L = 1.17 M. then 1.17 x 53.4g = 62.65g

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