H3PO4 ----------------> H2PO4- + H+ 0.30 0 0 0.30 - x x x Ka1 = [x][x]/[0.3-x] 6
ID: 699583 • Letter: H
Question
H3PO4 ----------------> H2PO4- + H+
0.30 0 0
0.30 - x x x
Ka1 = [x][x]/[0.3-x]
6.92 x 10^-3 = x^2/(0.30-x)
x = 0.0422
[H3PO4] = 0.30 - 0.0422 = 0.258 M
[H3PO4] = 0.258 M
[H2PO4-] = x
[H2PO4-] = 0.0422 M
[H+] = x = 0.0422 M
almost maximum H+ ions comes from first ionisation so
pH = -log [H+] = -log(0.0422)
pH= 1.38
[OH-] = kw/[H+] = 1 x10^-14/0.0422
[OH-] = 2.40 x10^-13 M
H2PO4- -----------------> HPO4^2- + H+
0.0339 0 0.0339
0.0339 -y y 0.0339 +y
Ka2 =[HPO4^2][ H+]/[H2PO4-]
6.2 x10^-8 = (y)(0.0339 +y)/(0.0339 -y)
by solving this y = 6.2 x10^-8
second ionisation constant value always =[HPO4^-2]
[H2PO4-]= 0.029-y = 0.029-6.2 x10^-8
[HPO4^-2] =6.2 x10^-8 M
HPO4^2 ---------------------> PO4^-3 + H+
6.2 x10^-8 0 0.0422
6.2 x10^-8- z z 0.0422 +z
Ka3 = [ PO4^-3][H+]/[HPO4^2]
4.7 x 10^-13 = (z)x(0.0422 +z)/ (6.2 x10^-8- z )
z= 6.91 x 10^-19 M
[PO4^-3] = z= 6.91 x 10^-19 M
Explanation / Answer
Phosphoric acid, HsPO(aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. the concentrations of all species in a 0.300 M phosphoric acid solution. Estimate the pH, and 2.16 721 12.32 Number Number Number H,PO Number HPOM Number Number
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