Question

The following solutions are mixed : 500 mL of 0.01 M AgNO, and 500 mL solution that was both
0.01 M in NaCl and 0.01 M in NaBr. Given K AgCl = 10-10, KAgBr = 5x10". Calculate the [Crº ) in the equilibrium solution​

Answer 1

Answer:

tion (equal volume) and if there were no precipitation. [NOΘ3]=[Ag⊕]=[CIΘ]=[BrΘ]=0.012=0.005M=5.0×10−3M.

AgBr is the more soluble salt (less Ksp means more soluble)and would take precendence in the precipitating reaction.

ii. Assume AgCI does not precipitate. In this case Ag⊕ and BrΘ would be removed by precipitation and the concentration of these two ions in solution would remain eqial to each other.

iii. [Ag⊕][BrΘ]=KspAgBr−−−−−−−−√=(5.0×10−13)12=7.1×10−7M

iv. Qsp or I.P. of AgCI=[Ag⊕][CIΘ]

=(7.1×10−7)(5.0×10−3M)=3.5×10−9

Qsp of AgCI>Ksp of AgCI. ( ∵ Some AgCI most also precipitate)

Hence the assumption in (ii) is wrong.

v. Since both holides precipitate, it is a case of simultaneous solubilities.

vi. By electronutrality

[Na⊕]+[Ag⊕]=[CIΘ]+[BrΘ]+[NOΘ3]

0.01+[Ag⊕]=[CIΘ]+[BrΘ]=0.005

or [CIΘ]+[BrΘ]−[Ag⊕]=0.005...(1)

vii. [Ag⊕][CIΘ]=10−10...(2)

[Ag⊕][BrΘ]=5×10−13..(3)

viii. Divide (2) by (3), [CIΘ][BrΘ]=200

This shows BrΘ plays a significant role in the total anion concentration of the solution.

xi. Moreover, [Ag⊕] must be negligible in (1) because of ionsolubility of two silver salts.

x. Therefore assume in (1) that [CIΘ]=0.005=5×10−3.

[BrΘ]=2.5×10−5

Check of assumption : Both [Ag⊕] and [BrΘ] are negligible compared with [CIΘ] of 0.005M