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What is the pH of 0.41 M H3PO4? The Ka values for phosphoric acid are Ka1 =...

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Chemistry

What is the pH of 0.41 MH3PO4? The Ka values forphosphoric acid are Ka1 = 7.5×10-3, Ka2 =6.2×10-8, and Ka3 = 3.6×10-13.

Answer & Explanation Solved by verified expert
4.0 Ratings (792 Votes)

Since Ka2 and Ka3 are very small as compared to Ka1
We will use only Ka1

H3PO4    -----> H +   H2PO4 -
0.41                      0           0       (initial)
0.41-x                  x           x        (at equilibrium)

Ka1 = [H+] [H2PO4-]/[H3PO4]
7.5*10^-3 = x*x / (0.41-x)

3.075*10^-3 - 7.5*10^-3 x = x^2
x^2 + 7.5*10^-3 x - 3.075*10^-3 = 0

solving for x, the positive value of x is 0.052

So,
[H+] = 0.052 M
pH = -log [H+]
       = -log (0.052)
       = 1.28
Answer: 1.28
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