Question

At 298 K, how many milligrams of silver
bromide can be dissolved in 20 litres of
water -

(Ksp - 5.0 * 10-13)
(Atomic wt. Ag = 108, Br=80)
(1) 2.66
(2) 3.66
(3) 4.66
(4) None of these

(Answer - 2.66)
​

Answer 1

Answer:

From the expression for Ksp, calculate the molar solubility of AgBr:

[Ag+] = [Br-] = x

5X10^-13 = x^2

x = 7.07 X10^-7 = molar solubility of AgBr

7.07 X 10^-7 mol/L X 20 L = 1.41X10^-5 mol AgBr X 187.8 g/mol = 2.66X10^-3 g X 1000 mg/g = 2.66 mg AgBr dissolved in 20 L H2O.

Answer 2

The correct answer is option 1.

Explanation:

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

                   s     s

Solubility product of AgBr = $K_{sp}=5.0\times 10^{-13}$

$K_{sp}=[Ag^+][Br^-]$

$5.0\times 10^{-13}=s\times s$

$s=\sqrt{5.0\times 10^{-13}}=7.0711\times 10^{-7} M$

Concentration of AgBr = Concentration of silver ions

( 1 mole of AgBr gives 1 mole of silver ions )

Concentration of AgBr = $7.0711\times 10^{-7} mol/L$

This means that $7.0711\times 10^{-7}$ moles of AgBr are dissolved in 1 L of solution.

Moles of AgBr in 20 L :

$7.0711\times 10^{-7} Mol/L\times 20 L=1.414\times 10^{-5} mol$

Mass of $1.414\times 10^{-5} mol$ of AgBr:

$1.414\times 10^{-5} mol\times 188 g/mol=0.00266 g$

0.00266 g = 2.66 mg (1 g =1000 mg)

The correct answer is option 1.

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