Answer to Question #158827 in General Chemistry for noga

Question

A20 ml of 0.02M solution of EDTA (Titrant) .28

was used to titrate 40 ml of solution containing

(A13+) aluminum ions of 0.01M . Calculate the

concentration of (A13+) ions at equilibrium? ( K eff

= 1.1 x 108).

A13+ + EDTA

* [A1 —EDTA]3

x 10-3 M 0.778

C)

x 10-5 M 7.78

x 10-7M 7.78

x 10-6 M 7.78

Answer 1

(40 / 1000) x 0.01 = 0.0004 mol Al³⁺ - initial

Al³⁺ + EDTA = Al-EDTA complex (1 : 1 ratio)

K = 1.1 x 10⁸

K =  = 1.1 x 10⁸

If 20 mL of EDTA was initially:

(20 / 1000) x 0.02 M = 0.0004 mol

Dilution: 40 + 20 = 60 mL

[Al³⁺] = [EDTA] = 0.0004 / 0.06 = 0.0067 M

x = 0.00669

0.0067 – x = 0.0067 – 0.00669 = 1 x 10^-5 M Al³⁺ at equilibrium