Question

What is the pH of a 0.041 M solution of Phosphorous acid (H2PHO3) in water? The value of Ka1 for H2PHO3 is 1.600 x 10-2 and the value of Ka2 is 7.000 x 10-7.

Answer 1

Let me be clear on this, 10-2 means 10^-2 right? aka
\[10^{-2}\]

right?

Answer 2

Oh what... there are 2 Kas in this?? What the hell... Not sure how to do this then.

Answer 3

So you need to find the moles of \(H^+\) generated from each dissociation:

\(H_2PO_3\rightarrow H^++HPO_3^-~~K_{a1}=1.6*10^{-2}\)

\(HPO_3^-\rightarrow H^++PO_3^{2-}~~K_{a2}=7*10^{-7}\)

by using two equilibrium expressions (one for each).

Start with the first one and use \([HPO_3^-]\) you get from it for the second.

Add the moles up from each dissociation and convert them to concentration i.e. molarity.

Finally, use \(pH=-log[H^+]\) to find the pH.

NOTE: assume a 1 L solution.