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Calculate the solubility of AgCl (Ksp = 1.8x10⁻¹⁰) at 25 degrees C in: a) pure water b) 3.0 M NH₃

User Cstack
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Final answer:

The solubility of AgCl in pure water at 25°C is 1.33 × 10-5 M. The solubility can be calculated using the solubility product constant (Ksp) equation. The presence of NH3 in a 3.0 M NH3 solution will affect the solubility of AgCl.

Step-by-step explanation:

The solubility of AgCl in pure water at 25°C is 1.33 × 10-5 M. This can be calculated using the solubility product constant (Ksp) equation, where Ksp = [Ag+][Cl-]. For AgCl, Ksp = 1.8 × 10-10, so by solving for x in the equation x2 = 1.8 × 10-10, we find that x = 1.33 × 10-5 M.

In a 3.0 M NH3 solution, the solubility of AgCl will change due to the presence of NH3 ions. However, this question only asks for the solubility in pure water, so the solubility in a 3.0 M NH3 solution is not relevant.

User Brian Hoover
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