titrate 25.00 mL of 0.1000 M I^- with 0.05000 M Ag^+

titrate 25.00 mL of 0.1000 M I^- with 0.05000 M Ag^+

 

 

What is [Ag^+] in solution once 10.00 mL of the Ag^+ titrant have been added?

 

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titrate 25.00 mL of 0.1000 M I^- with 0.05000 M Ag^+

I^-(aq) + Ag^+(aq) → AgI(s)

 

This is the reverse of the equilibrium:

AgI(s) ⇌ Ag^+(aq) + I^-(aq) ... Ksp = [Ag^+][I^-] = 8.3*10^-17

 

Because K for the titration reaction is large (1/Ksp = 1.2*10^16), products are favored, so we’d expect Ag^+ to be nearly used up during the titration until the equivalence point.

 

The expected equivalence volume of titrant Ve:

MV = M’V’

Ve = (0.1000)(25.00) / (0.05000) = 50.00 mL

 

 

What is [Ag^+] in solution once 10.00 mL of the Ag^+ titrant have been added?

 

> The initial mol of I^- was:

(0.1000)(25.00) = 2.500 mmol I^-

 

> The current mol of I^- is (1:1 rxn with Ag^+):

2.500 mmol - (0.05000)(10.00) = 2.000 mmol I^-

2.000 mmol / (25.00+10.00 mL) = 0.05714 M I^-

or

(40/50) (0.1000) (25.00/35.00) = 0.05714 M I^-

* (40/50) = (당량점까지 남은 부피/당량점 부피)

 

> Concentration of Ag^+ in equilibrium:

[Ag^+] = Ksp / [I^-] = (8.3*10^-17) / (0.05714)

= 1.453*10^-15 M Ag^+

 

 

답: 1.453*10^-15 M