Question

What is the molar solubility of agcl in 0. 30 m nacl at 25°C. ksp for agcl is 1. 77 × 10^-10.

Answer 1

Molar solubility of AgCl will be  0.59 ×  $10^{-10}$ M.

The amount of a chemical that can dissolve in one liter of a solution before reaching saturation is known as its molar solubility. This implies that the quantity of a substance it can disintegrate in a solution even before the solution becomes saturated with that particular substance is determined by its molar solubility.

A compound's molar solubility would be the measure of how many moles of such a compound must dissolve to produce one liter of saturated solution. The molar solubility unit will be mol L-1.

Calculation of molar solubility:

Given data:

M = 0.30 M

$K_{sp}$ = 1.77 × $10^{-10}$

The reaction can be written as:

AgCl ⇔ $Ag^{+} + Cl^{-}$

s            s         (s+0.30)

$K_{sp}$  = [$Ag^{+}$ ]+ [$Cl^{-}$]

1.77 × $10^{-10}$ = s (0.30)

s = 1.77 × $10^{-10}$  / 0.3

s = 0.59 ×  $10^{-10}$ M

Therefore, molar solubility of AgCl will be  0.59 ×  $10^{-10}$ M.

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