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

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