Question

Answer it properly..​

Answer 1

Question:

Ksp of AgNO₃ is 2 × 10⁻¹⁰. Calculate solubility in 0.1 M HNO₃?

Answer:

• Solubility (S) in 0.1 M HNO₃ is 2 × 10⁻⁹ moles/L

Given:

1. Ksp of AgNO₃ = 2 × 10⁻¹⁰.

Explanation:

$\rule{300}{1.5}$

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved where it Produces ions in solution. It is represented as Ksp.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

The Two Reactions we have:-

$\large\bigstar \: {\tt AgNO_3 \xleftrightharpoons{} Ag^+ + NO_3^-}$

$\large\bigstar \: {\tt HNO_3 \xleftrightharpoons{} H^+ + NO_3^-}$

Let the Solubility Product be "S".

Therefore, In given Medium (HNO₃) We can See Common- Ion Effect of NO₃⁻ ions.

Now, Its Given,

Molarity of solution as 0.1 M. Therefore, Solubility will be:-

• For Ag⁺ = S Moles/L.
• For NO₃⁻ = (S + 0.1) Moles/L.

Note:-

Here 0.1 is added to NO₃⁻ ions because of Common - ion Effect.

Now,

$\large\bigstar \: {\boxed{\tt Ksp = [Ag^+] \: [NO_3^-]}}$

$\large{\tt \hookrightarrow Ksp = [Ag^+] \: [NO_3^-]}$

Substituting the values,

$\large{\tt \hookrightarrow 2 \times 10^{-10} = [S] \: [S + 0.1]}$

• In the Case of NO₃⁻ ions "S" can Be neglected due as its Concentration will be very small due to Common ion Effect.

Therefore,

$\large{\tt \hookrightarrow 2 \times 10^{-10} = [S] \: [0.1]}$

$\large{\tt \hookrightarrow 2 \times 10^{-10} = [S] \: \Bigg[\dfrac{1}{10}\Bigg]}$

$\large{\tt \hookrightarrow 2 \times 10^{-10} \times 10 = [S] \: [1]}$

$\large{\tt \hookrightarrow 2 \times 10^{-10 + 1} = [S] \: [1]}$

$\large{\tt \hookrightarrow S = 2 \times 10^{-10 + 1}}$

$\huge{\boxed{\boxed{\tt S = 2 \times 10^{-9} \: Mole/L}}}$

∴ Solubility (S) in 0.1 M HNO₃ is 2 × 10⁻⁹ moles/L.

$\rule{300}{1.5}$