Chemistry, asked by PRACHISHETTY7596, 1 year ago

The solubility product of agbr is 4.9 * x10^-9 the solubility of agbr will be

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Answered by tallinn

Answer:- $7.0*10^-^5M$

Solution:- The ionization equation for AgBr is written as:

$AgBr(s)\rightarrow Ag^+(aq)+Br^-(aq)$

If the solubility of AgBr is s then the solubility for each product ion would also be same as there is 1:1 mol ratio. The solubility product, Ksp expression is written as:

$Ksp=[Ag^+][Br^-]$

Let's plug in the values in it:

$4.9*10^-^9=(s)(s)$

$4.9*10^-^9=(s)^2$

Taking square root to both sides:

$s=7.0*10^-^5$

So, the solubility of AgBr will be $s=7.0*10^-^5M$ .

avinashkumarpurushot: Wrong answer

tallinn: 7.0 *10^-5 M is also not wrong. The only thing is, M stands for molarity means what I have calculated is the molar solubility. If it wants it in terms of gram/Liter then we could multiply this by the molar mass of AgBr. Molar mass of AgBr is 187.77 g/mol. So, 7.0*10^-5 mol/L(187.77g/mol) = 1.3*10^-2 g/L

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