Enough of a monoprotic acid is dissolved in water to produce a 0.0136 M solution. The pH of the resulting solution is 2.45. Calculate the Ka for the acid.

Question

abb0t · Answer

Find the concentration first: 

[H+]: $$10^{-2.45}$$

anonymous · Answer

then???

abb0t · Answer

You know that for any monoprotic acid, it dissociates in the following manner: $$HA \rightleftharpoons H^+ + A^-$$

abb0t · Answer

Are you familiar with how to make an ICE table?

anonymous · Answer

[H+] = Ka - $$[H^+] = Ka - \frac{ \log [A] }{\log [B] ? }$$

anonymous · Answer

lol is the equation right? =))

anonymous · Answer

then find the -log of everything

anonymous · Answer

it results into this equation:

$$pH = pKa + \frac{ \log [B] }{ \log [A]? }$$

wherein A is the weak acid and B is the conjugate base

anonymous · Answer

to find the Ka, you have to derive the formula

anonymous · Answer

im not sure how to set up the ICE table..

anonymous · Answer

uh... i think that the info needed is incomplete? i'm not sure though O.O

anonymous · Answer

where did you get that question by the way? :))

anonymous · Answer

saplinglearning.com

anonymous · Answer

oh, okay. :)) ugh, i need to review Henderson-Hassellbalch equation @_@