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December 18, 2014

December 18, 2014

Posted by **Summer** on Sunday, January 22, 2012 at 12:11am.

What is the pH of a 0.089M solution of phosphorous acid (H2PHO3) in water? The value of Ka1 for H2PHO3 is 1.600 E-2 and the value of Ka2 is 7.000 E-7.

H2PHO3 + HOH --> HPHO3^-1 + HeO^+1

0.089M............0..........0

-x...............x..........x

.089-x...........x..........x

1.600E-2=x^2/.089-x

I used the 5% rule, so x= 3.774E-2

and then since Ka2 is so small, is it insignificant? I thought it was, so I went ahead and calculated pH.

pH=-Log(3.774E-2)

pH=1.42

- Chemistry -
**DrBob222**, Sunday, January 22, 2012 at 12:30amI don't think you can assume 0.089-x = 0.089. I think you must solve the quadratic (or solve it by successive approximations (iteration). I solved the quadratic and obtained about 0.0305. If I plug that in for x in

k1 = (x)(x)/(0.089-x) I get 0.0159 which is close. pH then is about 1.52 or so. Check my work. It's late here.

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