Question

How many grams of Agl (Ksp = 1.8 x 10-19) can be dissolved in 500.0 mL of pure water? (MM of AgCl = 143.35 g/mol) 1.3.0*10-1 2.3.2 x 10-3 3.9.6 x 10-4 4. 1.8 x 10-7

Answer 1

In this question AgI is given but molar mass of AgCl is given. I am using this molar mass only.

AgI$\rightleftharpoons$ Ag^+.  +. I^{​​​​​-}

K_{​​​​​​sp} = [Ag⁺] [I^{​​​​​​-} ] = 1.8 * 10^-10

^{​​​​​​}let molar concentration of ions be s M

K_{​​​​sp} = s * s = 1.8 * 10^-10

s = $\sqrt{}$ 1.8 * 10^-5 = 1.34 * 10^-5 M

1.34 * 10^-5 M means 1.34 * 10^-5 moles are present in 1 litre

1 litre or 1000 ml contains 1.34 * 10^-5 moles of AgI

For 500 ml, no of moles = 1.34 * 10^-5 * 500/ 1000 = 6.708 * 10^-6 moles

Number of moles = mass taken / molar mass

6.7 * 10^-6 = mass taken / 143.35

Mass taken = 143.35 * 6.7 * 10^-6 = 9.6 * 10^-4 g