Calculate the molar solubility of silver chloride in 0.15 m sodium chloride.
ksp of silver chloride = 1.6x10-10.

S = 1.1 × 10⁻⁹ M

Explanation:

NaCl is a strong electrolyte that dissociates according to the following expression.

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.

We can find the molar solubility (S) of AgCl using an ICE chart.

      AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I                         0            0.15

C                      +S              +S

E                        S            0.15+S

The solubility product (Ksp) is:

Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)

If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

c. electrical charges of ionic bonds.