Question

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium chloride (NaCl) and 0.05000 M potassium iodide (KI), what is the pAg of the solution after 15.00 mL of AgNO3 is added to the solution? Ksp, AgCl (s) = 1.82 x 10-10; Ksp, AgI(s) = 8.3*10-17.

b. Same titration as in (a), what is the pAg of the solution after 25.00 mL of AgNO3 is added to the above solution?

c. Same titration as in (a), what is the concentration of the Cl- ion after 85.00 mL of AgNO3 is added to the solution?

Answer 1

Ksp ( Ag Cl ) = 1.82 x 10^-10

[ Ag+] [Cl-] = 1.82 x 10^-10  

[ Ag⁺ ] = [ Cl^- ] = 1.35 x 10^-5

Similarly, Ksp ( Ag I ) = 8.3 x 10^-17

so [ Ag+ ] = [ I- ] = 0.91 x 10^-9

......... ..............or, = 9.1 x 10^-10   

Now assume that x ml of 0.100M NaCl is present out of 25 ml of the mixture solution ,so volume of KI solution = ( 25- x ) ml

& Concentration of Cl- = 0.1 x

.............................[ I- ] = ( 25 - x ) 0.05

Thus 0.1x + (25 - x ) (0.05 ) = 15 x 0.05

& x = 11.1 ml (which represents the volume of 0.1 M NaCl solution)

volume of KI solution =( 25 - 11.1) = 13.9 ml.

Check this concentration of [ I- ] to cause precipitation of Ag + with respect to Ksp ( Ag I ) = [ (13.9 x 0.05 ) ................................................................................................................................................../1000 ]

....................................................................................................................................... = 6.9 x 10^-4

[ Ag + ] in 15 ml of Ag NO3 = 15 x 0.05 / 1000 = 7.5 x 10^-4

so ionic product [ Ag + ] [ I - ] = ( 6.9 x !0^-4 ) ( 7.5 x 10^-4 ) = 5.18 x 10^-7 ,

this will cause precipitation of all I- , because the above product has exceeded the solubility product of

AgI and excess silver ion will remain in solution.

Concentration of excess Ag+ =   { ( 8.3 x 10^-7 ) / (6.9 x 10^-4 ) } - (7,5 x 10^-4 )

........................................... = ( 1.2 x 10^-3 ) - ( 7.5 x 10^-4 )

............................................ = 4.5 x 10^-4

b ) Questions on part b) & c) can also be attempted the same way , taking into consideration of Ksp ( AgCl ) only as I- have completly been precipitated by Ag NO3 used in question as in part a).

Glad to help. Please post remaining questions at partt b) & c) for detailled answers separately as fresh questions.