Calculate the solubility product constant, Ksp, of?
moles PbCl2 = 1.0 g/ 278.1 g/mol=0.0036 concentration of dissolved PbCl2 = 0.0036 mol/ 0.100 = 0.036 M [Pb2+]= 0.036 [Cl-]= 0.036 x 2 =0.072 M Ksp = [Pb2+][Cl-]^2 = 0.036 x (0.072)^2 =1.9 x 10^-4
Why is the solubility product of AgCl>AgBr>AgI?
The molar solubilities are given by root of Ksp for a 1:1 electrolyte, and consequentlythey decrease in the same order as the solubility products: AgCl>AgBr>AgI.This order is predicted by Fajan’s rules since the anions increase in size asCl-
Solubility product of AgCl is 4*10^-10 at 298 K. Solubility of AgCl in 0.04 M CaCl2 is?
[math]AgCl (s)[/math] disassociates in water in the following equation:[math]AgCl (s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)[/math]That means, [math]K_{sp} = [Ag^+(aq)][Cl^-(aq)][/math][math]4 \times 10^{-10} = [Ag^+(aq)][Cl^-(aq)] [/math]Since the same number of ions are produced, you could represent it using 'x':[math]4 \times 10^{-10} = [x][x][/math][math]x^2 = 4 \times 10^{-10}[/math][math]x = 2 \times 10^{-5}\text{ mol }dm^{-3}[/math]Hence, your solubility would be [math]2 \times 10^{-5}\text{ mol }dm^{-3}[/math].However, addition of a salt containing a common ion results in something called a 'common ion effect' which effectively reduces the solubility of the salt. This is because the addition of the same ion into the solution shifts the equilibrium backwards, creating more solid and less ions.[math]CaCl_2[/math] (aq) could be thought up to disassociate as:[math]CaCl_2 (aq) \rightarrow Ca^{2+} (aq) + 2 Cl^-(aq) [/math]So, [math]0.04[/math] mol of [math]CaCl_2 (s)[/math] can produce [math]0.08[/math] mol of [math]Cl^-(aq) [/math]ions. Now, I'll assume one thing: when you're mixing the two the volumes are equal to each other. this essentially halves the concentration so the new concentration of [math]Cl^-(aq)[/math] would be:[math]\frac{0.08 + 2 \times 10^{-5}\text{ mol }dm^{-3}}{2}[/math][math]= 0.04\text{ mol }dm^{-3}[/math]Now,[math]K_{sp} = [Ag^+(aq)][Cl^-(aq)][/math][math]4 \times 10^{-10} = [Ag^+(aq)][0.04][/math][math][Ag^+(aq)] = 1 \times 10^{-8} \text{ mol }dm^{-3}[/math]Note: [math]AgCl[/math] actually has a [math]K_{sp}[/math] of [math]2.0 \times 10^{-10}[/math]. But since the questions states another value, I will be using that instead.
The molar solubility of PbI2 is 1.5 x 10^-3 mol/L. What is the molar concentration of the iodide ion in saturated PbI2 solution?
In internet I saw: 1.4 x 10–6 mol/LPbI2 = [Pb] + 2 [I]-Ks = 1.4 x 10–6 = S x S2S = 0.0112 mol/L2S = 2 x 0.0112 = 0.0224 mol/L
Assuming that the solubility of PbCl2(s) is 1.6 x 10-2 mol/L at 25C, calculate the Ksp for this salt?
If solubility of PbCl2 is 1.6 x 10^-2 mol/L, then [Pb++] = 1.6 x 10^-2, and [Cl-] = 3.2 x 10^-2. Ksp = [Pb++][Cl-]^2 Ksp = (1.6x10^-2)(3.2x10^-2)^2 Ksp = (1.6x10^-2)(10.2x10^-4) Ksp = 1.6 x 10^-5
Calculate the concentrations of Pb2+ and Cl - at equilibrium?
moles Pb2+ = 0.0500 L x 0.25 M = 0.013 moles Cl- = 0.0500 L x 2.2 M = 0.11 total volume = 0.100 L [Pb2+]= 0.013 / 0.100 = 0.13 M [Cl-] = 0.11/ 0.100 = 1.1 M The precipitation reaction Pb2+ + 2Cl- >> PbCl2 requires a 1 - to - 2 ratio so Cl- is in excess Let us assume that the excess Cl- drives all Pb2+ out of solution. This means that 0.13 M Pb2+ reacts with 2 x 0.13 = 0.26 M of Cl- to produce PbCl2 leaving 1.1 - 0.26 = 0.84 M Cl- in excess. Let us now as much PbCl2 dissolve as needed to estabilish the PbCl2 saturation equilibrium let us say , x mol/L. this will rise the concentration of Cl- to 0.84 + 2x and will make the Pb2+ concentration equal to x. PbCl2 <-----> Pb2+ + 2Cl- [Pb2+]= x and [Cl-] = 2x + 0.84 Ksp = 1.6 x 10^-5 = (x)( 2x + 0.84)^2 x = 2.3 x 10^-5 [Pb2+] = 2.3 x 10^-5 M and [Cl-] = (2 x 2.3 x 10^-5) + 0.84= 0.84 M
How important is it to predict the solubility of an ionic crystal in a liquid?
Pretty damn important if you were about to die, and a leperchaun showed up and gave you a magic ionic crystal powder that you had to dissolve in a particular solvent and then drink the resulting elixir for your life to be saved. But the sneaky leperchaun didn’t specify which solvent to use to make the elixir…Let’s say you put the powder into a glass of water, and then discovered that the powder does not dissolve in water.You might well end up dead! Oh No!However, if you knew that the magic life-saving ionic crystal powder was completely soluble in ethyl alcohol, then you would be able to drop the powder into a shot of whiskey, dissolve it and drink the whole elixir down, and your life would have been saved, and you would have been overjoyed that a leperchaun not only saved your life but also made you a little drun…haha!So, solubility of an ionic crystal in a given solvent is an important parameter to predict….
What is the solubility?
Ksp = [Pb++][Cl-]² For every PbCl2 that dissolves, one Pb++ and two Cl- are in solution. So, we can let x = molar solubility Ksp = 1.6x10^-5 = (x)(2x)² But since there already is 0.15 M of Cl- (from the HCl), then 1.6x10^-5 = (x)(2x + 0.15)² To solve this, we assume 0.15>>2x, so we can ignore the 2x in the equation: 1.6x10^-5 = (x)(0.15)² Solving, x = 7.1x10^-4, The answer is D).
The solubility of lead (II) chloride (PbCl2) is?
The equilibrium is PbCl2 (s) <=> Pb2+ (aq) + 2 Cl- (aq) the requirement for equilibrium is given by Ksp = [Pb2+][Cl-]^2 we are told that the concentrations are [Pb2+] = 1.6 x 10^-2 M conforming to the chemical equation [Cl-]= 2 x 1.6 x 10^-2 = 3.2 x 10^-2 Ksp = ( 1.6 x 10^-2)(3.2 x 10^-2)^2 =1.6 x 10^-5