Chemistry
posted by Summer .
I'm doing online chemistry homework, and the site keeps telling me I have the wrong answer, but I'm not sure what I'm doing wrong. please help?
What is the pH of a 0.089M solution of phosphorous acid (H2PHO3) in water? The value of Ka1 for H2PHO3 is 1.600 E2 and the value of Ka2 is 7.000 E7.
H2PHO3 + HOH > HPHO3^1 + HeO^+1
0.089M............0..........0
x...............x..........x
.089x...........x..........x
1.600E2=x^2/.089x
I used the 5% rule, so x= 3.774E2
and then since Ka2 is so small, is it insignificant? I thought it was, so I went ahead and calculated pH.
pH=Log(3.774E2)
pH=1.42

I don't think you can assume 0.089x = 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)/(0.089x) I get 0.0159 which is close. pH then is about 1.52 or so. Check my work. It's late here.