Determining the Ksp of Calcium Hydroxide Pre Lab Questions Determine the Ksp of
ID: 636520 • Letter: D
Question
Determining the Ksp of Calcium Hydroxide Pre Lab Questions Determine the Ksp of calcium fluoride (CaF2), given a molar solubility of 4.58 x 105 mol/L 1. 2. Determine the molar solubility of potassium chromate (KaCrO4) given a Ksp of 142. 3. When a common ion is introduced, should the molar solubility of a molecule increase, or decrease? Why? Discussion Questions 1. Based on your results, could you predict the solubility of a calcium hydroxide solution following the addition of a known amount of calcium chloride? Why or why nota What other factors besides a common ion, can affect solubility? Design an experiment to determine molar solubility that involves changing a different variable. 2.Explanation / Answer
Answer of pre lab questions
1. To find Ksp of CaF2 :
Here are simple steps to find out:
(a) Firstly break the compound into it’s successive ions and balance them.
CaF2 = Ca2+ + 2F-
(b) now for finding Ksp we have to use a symbol to determine molar solubility so, here we are using an ‘s’ (known as solubilty) . And write it just below the ions Ca2+ and 2F-.
CaF2 = Ca2+ + 2F-
Ksp =. s +. (2s)2
NOTE: while finding Ksp when you distribute ‘s’ below ions then for an ion with a coefficient ( in this case we have flourine with a coefficient) you have to wrte the coefficient and also raise it as power. Like in this case flourine with a coefficient ‘2’ we wrote 2 as coefficient of ‘s’ and also raised it as power. But if the coefficient would have been say, 3 then we would have written 3s3.
(c) now proceed with basic calculations like this,
Ksp = S (for Ca2+) + (2S)2 ( for 2F-) and now put the value of S which is 4.58 * 10-5
We get Ksp= 4S3 ( after squaring and adding) and now put the value of S
After multiplying we will get Ksp= 18.32 * 10-5.
2. For molar solubilty of K2CrO4
Repeat the same steps as we did to find out Ksp except the difference we are given the value of Ksp and we have to find ‘S’.
K2CrO4 = 2K+ + CrO4- So, the next equation will be Ksp = 2S2 + S , after solving we get Ksp = 4S3
Then solve as you normally do, putting the value of Ksp .
142 = 4S3 then after dividing and cubing we get S= 3.28.
3. Yes , after addition of common ion of a salt ,the solubilty of a salt is reduced in a solution which already contains an ion which is common with that salt. The reason being that addition of a salt in a solution with a common ion precipitates the the ion with lower solubility. Lets simplify this with an example.
Say you have a glass of impure water which has silver chloride as an impurity. Now to make it safe for drinking you poured some calcium chloride with chlorine as a common ion. You will see that because silver has lower solubilty as compared to calcium it will get precipitated at the bottom of glass. This effect is known to us by ‘ the common ion effect’.
4. When we add calcium chloride to a solution containing calcium hydroxide. We should know that chlorine ion is more electronegative than hydroxide ion so chlorine has higher tendency to get precipitated in the solution. So final resulting soltion should be calcium hydroxide and chloride will be the precipitate.
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