Question

I'll get a question like the one below and my professor says that we need to find all three equivalence points in order to know what stage we are in. I know how to find the first equivalence point, but I don't know how to find the 2nd and 3rd equivalence point.

Please show me how to calculate all equivalence points for the problem below. Thanks for your help!!!

20.0mL of 0.050M phosphoric acid is titrated with 0.100M NaOH.(Ka1=7.11 x10-3, Ka2=6.32x10-8, Ka3=4.5x10-13)

Answer 1

(a) First equivalence point

moles of NaOH = moles of H3PO4

0.05 M x 0.02 L = 1 x 10^-3 mols is mols of base added

Volume of base added = 1 x 10^-3 mols/0.1 M = 0.01 L

[H2PO4-] = 1 x 10^-3/0.03 = 0.033 M

salt hydrolyzes as,

H2PO4- + H2O <==> H3PO4 + OH-

let x amount hydrolysed

Kb1 = Kw/Ka1 = 1 x 10^-14/7.11 x 10^-3 = [H3PO4][OH-]/[H2PO4-] = x^2/0.033

x = [OH-] = 2.15 x 10^-7 M

pOH = -log[OH-] = 6.67

pH = 14 - pOH = 7.33

(b) second equivalence point

moles of base = 2 x moles of acid = 2 x 1 x 10^-3 = 2 x 10^-3 mols

Volume of base added = 2 x 10^-3/0.1 = 0.02 L

[HPO4-] = 2 x 10^-3/0.04 = 0.05 M

salt hydrolyzes as,

HPO4^2- + H2O <==> H2PO4- + OH-

let x amount hydrolysed

Kb2 = Kw/Ka2 = 1 x 10^-14/6.32 x 10^-8 = [H2PO4-][OH-]/[HPO4^2-] = x^2/0.05

x = [OH-] = 8.89 x 10^-5 M

pOH = 4.05

pH = 9.95

(c) Third equivalence point

moles of base = 3 moles of acid = 3 x 1 x 10^-3 = 3 x 10^-3 mols

Volume of base added = 3 x 10^-3/0.1 = 0.03 L

[HPO4-] = 3 x 10^-3/0.05 = 0.06 M

salt hydrolyzes as,

PO4^3- + H2O <==> HPO4^2- + OH-

let x amount hydrolysed

Kb3 = Kw/Ka3 = 1 x 10^-14/4.5 x 10^-13 = [HPO4^2-][OH-]/[PO4^3-] = x^2/0.06

x = [OH-] = 0.036 M

pOH = 1.44

pH = 12.56