Question

A 25.00 mL solution of 0.08680 M NaI is titrated with 0.05120 M AgNO3 . Calculate pAg+ following the addition of the given volumes of AgNO3 . The ?sp of AgI is 8.3×10^−17

36.00 mLpAg+ =   ?

?e pAg+ = ?

47.40 mL pAg+ = ?

Answer 1

Consider a titration reaction as AgNO_3(aq) +NaI _(aq) AgI _(s) + NaNO_3(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⁺    0.08680 M  x 25.00 ml / 0.05120 M

V _Ag⁺ 42.38 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 ) ( 0.08680 mol/L) - (0.036 L ) ( 0.05120 mol/L) ( No. of moles = Volume in L x Molarity )

= 0.00217 mol  - 0.001843 mol

=0.000327 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 = 0.000327 mol /0.061 L = 0.00536 M

[ Ag⁺ ]=  Ksp / [Br^-]

[ Ag⁺ ] = 8.3 x 10^-17 /0.00536

[ Ag⁺ ] = 1.55 x 10^-14 M

p Ag⁺ = -log  [ Ag⁺ ] = - log 1.55 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 47.40 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 ⁺= (0.0474 L ) ( 0.05120 mol/L) - (0.025 L ) ( 0.08680 mol/L)

Moles of Ag ⁺= 0.00243 mol - 0.00217 mol = 0.000257 mol

Volume of solution after addition 47.40 ml of 0.05120 M Ag(NO₃ )  = (0.025 + 0.0474 ) = 0.0724 L

[Ag⁺ ]= No. of moles / Volume of solution in L = 0.000257 mol / 0.0724 L = 0.003549 M

p Ag ⁺ = -log [Ag ⁺ ]

p Ag ⁺ = -log(0.003549)

p Ag ⁺ =2.45