In: Chemistry
Phosphoric acid is a triprotic acid, so H3PO4, H2PO4 -, and HPO4 2- are all acids. Which of the solutions below would have the lowest pH?
0.1 M NaH2PO4
0.1 M Na2HPO4
0.1 M Na3PO4
answer : 0.1 M NaH2PO4 solution has lowest pH value
pH orders : 0.1 M NaH2PO4 < 0.1 M Na2HPO4 < 0.1 M Na3PO4
detailed solution :
Ka1 = 7.1 x 10-3 pKa1 = 2.15.
Ka2 = 6.3 x 10-8 , pKa 2 = 7.20.
Ka3 = 4.5 x 10-13 , pKa3 = 12.35
(a) 0.1 M NaH2PO4
H2PO4- -------------------> HPO4-2 + H+
0.1 0 0 ---------------> initial
0.1-x x x
Ka2 = [HPO4-2][H+]/[H2PO4-]
Ka2 = x^2 / (0.1-x)
6.3 x 10-8 = x^2 / (0.1-x)
x^2 + 6.3 x 10^-8 x - 6.3 x 10^-9= 0
x = 7.93 x 10^-5
[H+] = x = 7.93 x 10^-5 M
pH = -log[H+]
pH = -log (7.93 x 10^-5 )
pH = 4.1-------------------------------------------(1)
(b) 0.1 M Na2HPO4
similar solution
HPO4-2 ---------------------> H+ + PO4-3
0.1-x x x
Ka3 = x^2 / 0.1-x
4.5 x 10^-13 = x^2 / 0.1-x
x^2 + 4.5 x 10^-13 x - 4.5 x 10^-14 = 0
x = 2.12 x 10^-7
x = [H+]=2.12 x 10^-7 M
pH = 6.67 --------------------------------------------(2)
(c) 0.1 M Na3PO4
PO4-3 + H2O ---------------------> HPO4-2 + OH-
0.1 0 0 --------------> initila
0.1-x x x
Kb = x^2 / 0.1-x
Kw / Ka3 = x^2 / 0.1-x
1.0 x 10^-14 / 4.5 x 10^-13 = x^2 / 0.1-x
0.022 = x^2 / 0.1-x
x^2 + 0.022 - 2.22 x 10^-3 = 0
x = 0.0037
x = [OH-] = 0.0037 M
pOH = -log[OH-]
pOH = -log (0.0037)
pOH = 2.43
pH + pOH = 14
pH = 14 -pOH
pH = 11.57 --------------------------------------------------------(3)