Question
please help!
For HyPO4, the Kai is so large, the "x is small approximation" is not valid. Asa 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 (Ke2 and Ka) are so small that the approximation works for in these cases. (a) Calculate the pH for the 0.05M H,PO, solution. Use the quadratic formula. Use the quadratic formula to calculate the pH at the halfway point (first half- Equiv. point) for HjPO.. Assume that the initial molarity of the H,POs 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 H:POs and H2PO.construct an ICE, and use the equations below to calculate the pH. (b) becomes Kal= (x) ( [H.PQ1+x) IH,PO,]-x) Kal =
Explanation / Answer
we know that pH = -log [H+]
In H3PO4 three H+ ions are present and [H+] = 0.05 M
therefore, pH = -log [H+]
pH = - log[3*0.05] = -log[0.15] = 0.82
pH = 0.82 for H3PO4
