In: Chemistry
A 25.0025.00 mL solution of 0.085700.08570 M NaINaI is titrated with 0.050300.05030 M AgNO3AgNO3. Calculate pAg+pAg+ following the addition of the given volumes of AgNO3AgNO3. The ?spKsp of AgIAgI is 8.3×10−1
36.00 mL pAg+ =___________________________
Ve pAg+ =______________________________
48.00 ml pAg+____________________________
I think there is minor mistake in Ksp value of AgI which should be 8.3 x 10-17
Consider a titration reaction as AgNO3(aq) +NaI (aq) AgI (s) + NaNO3(aq)
Calculation of equivalence point
We have, M Ag+ x V Ag+ M I-x V I-
V Ag+ M Br-x V I- / M Ag+
V Ag+ 0.08570 M x 25.00 ml / 0.050300 M
V Ag+ 42.59 ml
Consider dissolution of AgI as AgI (s) Ag+(aq) + I -(aq)
Ksp =[ Ag+ ][I-]=8.3 x 10-17
[ Ag+ ]= Ksp / [I-]
1) pAg + when 36.00 ml Silver nitrate added
Before equivalence point , all added Ag+are consumed by I - and excess I - remains in the solution . We can calculate concentration of non reacted I - as
Moles of I - = original moles of I - - moles of Ag+ added
= (0.025 L ) ( 0.08570 mol/L) - (0.036 L ) ( 0.05030 mol/L) ( No. of moles = Volume in L x Molarity )
= 0.0021425 mol - 0.0018108 mol
=0.0003317 mol
Volume after addition of 36.00 ml AgNO3 = (0.025+0.036)= 0.061 L
[I-] = No. of moles/ Volume of solution in L = 0.0003317 mol /0.061 L = 0.005438 M
[ Ag+ ]= Ksp / [I-]
[ Ag+ ] = 8.3 x 10-17 /0.005438
[ Ag+ ] = 1.526 x 10-14 M
p Ag+ = -log [ Ag+ ] = - log 1.526 x 10-14 = 13.8
2) pAg + at equivalence point
At equivalence point all Ag + is consumed by I -. But there is equilibrium between solid AgI and dissolved ions. Hence , at equilibrium the concentration of Ag + ions will be due to dissociation of solid AgI.
Consider dissolution of AgI as AgI (s) Ag +(aq) + I -(aq)
For above reaction, Ksp =[Ag +][I - ]=8.3 10-17
If S is the solubility of AgI then [ Ag +] = [I - ]= S
Therefore, S S = 8.3 10-17
S 2 = 8.3 10-17
S = 9.11 10-09 M =[ Ag +] = [I - ]
We have, p Ag + = -log [ Ag + ]
p Ag + = -log( 9.11 10-09 )
p Ag + = 8.04
3) pAg + when 48.0 ml Silver nitrate added
After equivalence point, there is excess of Ag + ions in the solution.
Moles of Ag += added moles of Ag + - moles of I - added
Moles of Ag += (0.048 L ) ( 0.05030 mol/L) - (0.025 L ) ( 0.08570 mol/L)
Moles of Ag += 0.0024144 mol - 0.0021425 mol = 0.0002719 mol
Volume of solution after addition 48 ml of 0.05120 M Ag(NO3 ) = (0.025 + 0.048 ) = 0.073 L
[Ag+ ]= No. of moles / Volume of solution in L = 0.0002719 mol / 0.073 L = 0.0037247 M
p Ag + = -log [Ag + ]
p Ag + = -log(0.0037246)
p Ag + =2.43
Your one thumbs up will help me lot. Thanks!