Question

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08160 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volumes of AgNO3 are added:

(a) at 35.10 mL

(b) at Ve (volume at equilibrium)

(c) at 47.10 mL

Answer 1

  

   

    

Answer 2

Thereaction between Naland AgNO3 is asfollows Nal (aq) +AgNO (aq)Ag(s) +NaNO3(aq) Dissociation of AgIasfollows Ag (s)Ag (aa)+I(aą) K[Ag ]-8.3x10 a) Calculate concentrati on of Iremaining after addition of 35.10mL of AgNO3 asfollows 25.00mL x0.08160 M-35.10mLx0.05190 M (25.00 +35.10)mL 0.003632M Calculate concentrati on of Ag as follows: [Ag+]= 8.3x1017 0.003632 M -2.2850 × 10-14 M CalculatepAg as follows pAg-log[Ag ] log(2.2850x10-" M) 13.64 Therefore,pAgis 13.64 b) At equlibrium volume, [Ag+ ] = [lj Calculate concentrati on of Ag asfollows = [Ag+ ][Ag+ ] V8.3×10-17 9.110x109 M CalculatepAg as follows pAg log[Ag] log(2.2850x10-" M) 8.040 Therefore,pAgis8.040 c) Calculate concentrati on of Ag* remaining after addition of 47.10mL of AgNO as follows [Ag - 47. 10mL × 0.05190M-25.00 mL × 0·08160M (25.00 +47.10)mL 0.005610M CalculatepAg as follows pAg-log[Ag ] log(0.005610 M) =2.251 Therefore,pAgis2.251