millimols HPO4^- = mL x M = approx 14 but you need to use a better number than that so recalculate all of these.
mmols HCl added = 50 x 0.275 = approx 14.
.......HPO4^2- + H^+ ==> H2PO4^-
I......14........0........0
add .............14............
C.....-14.......-14......+14
E......0..........0.......14
So you have prepared 14 mmols H2PO4^- in 100 mL so the pH will be determined by the 0.14M acid (that's 0.275/2 = ?).
......H2PO4^- + H2O ==> OH^- + H3PO4
I.....0.14...............0.......0
C......-x................x.......x
E.....0.14-x.............x.......x
Kb for H2PO4^- = (Kw/Ka for H3PO4) = (x)(x)/(0.14-x)
Solve for x = (OH^-) and convert to pH. Post your work if you get stuck.
Estimate the pH of the solution that results when this added to 50 mL of 0.275 M Na2HPO4(aq)
- 50 mL of 0.275 M HCl(aq)
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