AgI 침전적정. 0.1000 M I^- 25.0 mL + 0.0500 M Ag^+

AgI 침전적정. 0.1000 M I^- 25.0 mL + 0.0500 M Ag^+

 

 

Consider the titration of 25.0 mL 0.1000 M iodine ion I^-

with 0.0500 M silver ion Ag^+, which makes the silver iodide precipitate:

Ag^+(aq) + I^-(aq) → AgI(s)

 

The reverse of the dissolution of is

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

 

a. What volume of Ag^+ titrant is needed to reach the equivalence point?

b. What will be the concentration of silver ion, [Ag^+], and iodine [I-] if you just add 1.0 mL less Ag before reaching the equivalence point?

c. What will be pAg^+, if you just add 1.0 mL extra Ag after the equivalence point?

 

---------------------------------------------------

 

a. MV = M’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(0.1000 M) (25.0 mL) = (0.0500 M) (? mL)

? = (0.1000) (25.0) / (0.0500) = 50.0 mL

 

 

 

b. V’ = 49.0 mL

(0.0500 mmol/mL) × 49.0 mL = 2.45 mmol Ag^+

---> 이만큼 AgI(s) 생성.

---> 또한 이만큼 I^- 소모.

 

 

초기 I^-의 몰수를 계산하면,

(0.1000 mmol/mL) × 25.0 mL = 2.50 mmol KI

 

 

남아있는 [I^-]를 계산하면,

(2.50-2.45 mmol) / (25.0+49.0 mL)

= 0.000676 M I^-

 

 

Ksp = [Ag^+][I^-] 이므로,

[Ag^+] = Ksp / [I^-]

= (8.3×10^(-17)) / 0.000676

= 1.23×10^(-13) M

 

 

 

c. V’ = 51.0 mL

[Ag^+] = [(0.0500 mmol/mL) × 1.0 mL] / (25.0+51.0 mL)

= 0.000658 M

 

 

pAg^+ = -log[Ag^+]

= -log(0.000658)

= 3.18

 

 

 

[키워드] AgI 침전적정 기준문서