Question

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+...

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+____________________________

Solutions

Expert Solution

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!


Related Solutions

A 25.0−mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate...
A 25.0−mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution (a) 10.0 mL (b) 12.5 mL (c) 15.0 mL
A 26.0−mL solution of 0.120 M CH3COOH is titrated with a 0.220 M KOH solution. Calculate...
A 26.0−mL solution of 0.120 M CH3COOH is titrated with a 0.220 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.00 mL (b) 5.00 mL
A 10.0−mL solution of 0.570 M NH3 is titrated with a 0.190 M HCl solution. Calculate...
A 10.0−mL solution of 0.570 M NH3 is titrated with a 0.190 M HCl solution. Calculate the pH after the following additions of the HCl solution a)0mL b)10ml c)30mL d)40mL
A 10.0-ml solution of 0.780 M NH3 is titrated with a 0.260 M HCL solution. calculate...
A 10.0-ml solution of 0.780 M NH3 is titrated with a 0.260 M HCL solution. calculate the pH after the following additions of the hcl solution: A. 0.00 B. 10.0 C. 30.0 D. 40.0
A 50.0 mL solution of 0.167 M KOH is titrated with 0.334 M HCl . Calculate...
A 50.0 mL solution of 0.167 M KOH is titrated with 0.334 M HCl . Calculate the pH of the solution after the addition of each of the given amounts of HCl . 0.00 mL pH = 7.00 mL pH = 12.5 mL pH = 19.0 mL pH = 24.0 mL pH = 25.0 mL pH = 26.0 mL pH = 31.0 mL pH =
A 30.00 mL solution of 0.235 M CH2NH2 was titrated with 0.150 M H2SO4. a) Calculate...
A 30.00 mL solution of 0.235 M CH2NH2 was titrated with 0.150 M H2SO4. a) Calculate the pH of the 30.00 mL solution of 0.235 M CH2NH2 b) Calculate the volume of 0.150 M H2SO4 required to neutralize the 30 mL solution of 0.235 M CH2NH2. c) Calculate the volume of 0.150 M H2SO4 required to neutralize half the mole of CH2NH2 present 30 Ml solution of 0.235 M CH2NH2 d) Calculte the pH of the solution in (c). e)...
A 50.0 mL solution of 0.199 M KOH is titrated with 0.398 M HCl. Calculate the...
A 50.0 mL solution of 0.199 M KOH is titrated with 0.398 M HCl. Calculate the pH of the solution after the addition of the following amounts of HCl. a) 0.00 mL HCl b) 7.00 mL HCl c) 12.5 mL HCl d) 20.0 mL HCl
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the...
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added. 22.5 mL of the acid
A 83.0 mL sample of 0.0500 M HNO3 is titrated with 0.100 M KOH solution. Calculate...
A 83.0 mL sample of 0.0500 M HNO3 is titrated with 0.100 M KOH solution. Calculate the pH after the following volumes of base have been added. (a) 11.2 mL pH = _________ (b) 39.8 mL pH = __________    (c) 41.5 mL pH = ___________ (d) 41.9 mL pH = ____________    (e) 79.3 mL pH = _____________
A 71.0 mL sample of 0.0400 M HIO4 is titrated with 0.0800 M RbOH solution. Calculate...
A 71.0 mL sample of 0.0400 M HIO4 is titrated with 0.0800 M RbOH solution. Calculate the pH after the following volumes of base have been added. (a) 10.3 mL pH = (b) 34.4 mL pH = (c) 35.5 mL pH = (d) 36.2 mL pH = (e) 63.9 mL pH =
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT