4. For HPO, the Kal is so large, the \"x is small approximation\" is not valid.

Question

4. For HPO, the Kal is so large, the "x is small approximation" is not valid. As a result, pH is not equal to pKal at the first equivalence point. To more accurately determine the pH, an ICE table must be constructed and the quadratic formula should be used. The other K values (Ka2 and Kas) are so small that the approximation works for in these cases. (a) Calculate the pH for the 0.05M HyPO, solution. Use the quadratic formula Use the quadratic formula to calculate the pH at the halfway point (first half- Equiv. point) for HyPO. Assume that the initial molarity of the HPO4 is 0.0500 M (and that the volume is 25.00 mL) and that it takes 6.25 mL of 0.100 M NaOH to reach the first halfway point. Calculate the new molarities of HyPO, and HyPO.construct an ICE, and use the equations below to calculate the pH. (b) tri HPO-1 becomes K = T1,Po,J-X) [H PO,]

Explanation / Answer

a)

H3PO4 <----------------------> H+    + H2PO4-

0.05                                  0                0 -----------------> I

-x                                         +x                +x -----------------> C

0.05-x                             x                  x ------------------> E

Ka1   = x^2 /0.05 -x = 6.9 x 10^-3

x^2 + 6.9 x 10^-3 x - 3.45 x 10^-4 = 0

by solving this

x = 0.0154

[H+] = 0.0154 M

pH = -log [0.0154]

pH = 1.81

b)

at first half way point   [H3PO4] = [H2PO4-]   .so Ka 1 = H+ concentration

Ka = 6.9 x 10^-3 = [H+]

pH = 2.16

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