How does the solubility of AgClin pure water compare to its solubility in 0.15M NaCl?

Question

Ksp of AgCl= 1.6*10^-10

Answer 1

The NaCl is a freely soluble salt. A concentration of 0.15 M is far from saturated for NaCl, so no equilibrium will exist between solid NaCl and its ions. Rather, the entire solid mass of NaCl will completely dissociate in the solution, and only the ions will be present. If the amount of solid NaCl added to the water is enough to provide 0.15 moles of NaCl for each liter of solution, then the concentration of Na+ and Cl- in the solution will both be 0.15 M.
Since we will now be dissolving AgCl in this 0.15 M NaCl solution, the Cl- is a common ion. Its presence will suppress the solubility of AgCl, so it will be less soluble in this solution than it was in pure water. When the AgCl is dissolved, it establishes the usual equilibrium between solid AgCl and its ions in solution:

AgCl(s) <----------> Ag+(aq) + Cl-(aq)

The AgCl is the sole source of Ag+ ions, but Cl- will be present even before the AgCl dissolves due to the NaCl in the solution. In this system, only the Ag+ concentration is representative of the molar solubility of AgCl. The concentration of Cl- will not be indicative of the amount of AgCl that dissolved, since the majority of the Cl- ions will have come from NaCl, not AgCl.

KSP = [Ag+]eq [Cl-]eq = (X)(0.15 + X) = 1.6 x 10-10

Recall that the molar solubility of AgCl in distilled water was only 1.26 x 10-5 mol/L. Here, the molar solubility will be even lower, because of the common ion. The variable X will be very small in comparison to 0.15. That is,

0.15 + X ≈ 0.15

With this in mind, the equation

With this in mind, the equation

(x)(0.15 + X) = 1.6 x 10-10

reduces to

0.15X = 1.6 x 10-10

and dividing both sides by 0.15, we have

X = 1.6 x 10-10/ 0.15= 1.07 x 10-9


Therefore,

Molar solubility = [Ag+]eq = X = 1.07 x 10-9 mol/L

Comparing this to the molar solubility in pure water (1.26 x 10-5 mol/L), we see that the solubility has been reduced by a factor of more than 11000.

1.26x 10-5 mol/L
---------------- = 11.8 x 103
1.07 x 10-9 mol/L

Answer 2

very few issues do no longer dissolve in something. basically approximately each and every factor dissolves interior the water to a pair quantity. Water is seen easy solvent. They (NaCl crystals) are soluble interior the water. different solutions factors to(explains to a pair quantity) the chemistry and mechanism at the back of the solubility of a substance in a solvent.

Answer 3

The solubility in water is given by root Ksp.

The solubility in NaCl is given by Ksp/0.15.