PHYSICS HIGH SCHOOL

Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

Answers

Answer 1

Answer:

Answer : The maximum concentration of silver ion is $5* 10^(-12)m$

Solution : Given,

$K_(sp)$ for AgBr = $5* 10^(-13)$

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_(sp)=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_(sp)$ expression, we get

$5* 10^(-13)=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5* 10^(-12)m$

Therefore, the maximum concentration of silver ion is $5* 10^(-12)m$ .

Answer 2

Answer: [Br⁻¹] = [NaBr] = 0.1

Ksp = [Ag⁺] x [Br⁻]
5 x 10⁻¹³ = [Ag⁺] x 0.1
[Ag⁺] = 5 x 10⁻¹²

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