Calculate solubility of AgCl (in g/L) in a 6.5*10^-3 M silver Nitrate solution ksp= 1.8 * 10^-10
(AgCl molecular mass is 143.3g) any idea on how to do this this is what i tried>
AgCl--> Ag + Cl
I 6.5*10^-3 0
C same x
E same x
ksp = [Ag][Cl]
(1.8 * 10^-10) = (6.5 * 10^-3)x
x = 2.77 * 10^-8 M
I don't know what to do next please help me thank you
What you've done is almost correct assuming the spacing is ok. The numbers are but some are in the wrong place. This forum doesn't do spacing very well. I'll try to improve that.
.......... AgCl ==> Ag^+ + Cl^-
I.........solid.....0.......0
C.........solid.....x.......x
E.........solid.....x.......x
Then the AgNO3 ionizes completely as follows:
............AgNO3 ==> Ag^+ + NO3^-
E....................6.5E-3
Ksp = 1.8E-10 = (Ag^+)(Cl^-)
(Ag^) from AgCl is x
(Ag^+) from AgNO3 is 6.5E-3
Total is 6.5E-3 + x so the equation looks this way.
1.8E-10 = (6.5E-3+x)(x)
You have solved this the easy way by assuming 6.5E-3+x = 6.5E-3 and I think that probably is ok. You could go through the quadratic and see. At any rate this is the solubility of AgCl in mols/L. Convert that to grams/L by
grams AgCl = mols AgCl x molar mass AgCl.
Ah, the solubility of AgCl, a classic question in chemistry! It seems like you're on the right track with the initial setup. But fear not, for the Clown Bot is here to help you make sense of it all!
So, you've correctly set up the balanced equation for the dissolution of AgCl and written the initial concentrations of silver and chloride ions. Kudos to you! Now, all we need to do is solve for the solubility of AgCl in grams per liter (g/L).
To do that, we need to take into account the molar mass of AgCl. You mentioned it is 143.3g, which is great! Now, all we have to do is convert the solubility from molarity (mol/L) to grams per liter.
Here's how you can do it:
1. Calculate the molar solubility (in moles per liter) using the equation you set up: x mol/L (you found x = 2.77 * 10^-8 M).
2. Multiply the molar solubility (x) by the molar mass of AgCl (143.3g/mol) to get the solubility in grams per liter.
So, the solubility of AgCl in grams per liter (g/L) is:
(2.77 * 10^-8 M) * (143.3 g/mol) = 3.97 * 10^-6 g/L
There you have it! The solubility of AgCl in a 6.5 * 10^-3 M silver nitrate solution is approximately 3.97 * 10^-6 g/L. I hope that helps you clown around with solubility calculations!
To calculate the solubility of AgCl in g/L, you need to convert the concentration of AgCl in moles per liter to grams per liter.
The molar mass of AgCl is 143.3 g/mol.
First, convert the concentration to moles per liter:
Concentration of AgCl = 2.77 * 10^-8 mol/L
Next, convert the concentration to grams per liter:
Solubility of AgCl = (2.77 * 10^-8 mol/L) * (143.3 g/mol)
Solubility of AgCl = 3.97 * 10^-6 g/L
Therefore, the solubility of AgCl in a 6.5 * 10^-3 M silver nitrate solution is approximately 3.97 * 10^-6 g/L.
To calculate the solubility of AgCl in a 6.5 * 10^-3 M silver nitrate solution, you have correctly set up the equation for the dissociation of AgCl. However, you need to convert the concentration of AgCl from moles per liter (M) to grams per liter (g/L). Here's how you can proceed:
1. Determine the molar mass of AgCl:
Ag = 107.87 g/mol
Cl = 35.45 g/mol
AgCl = Ag + Cl = 107.87 + 35.45 = 143.32 g/mol
2. Calculate the molar solubility (x) of AgCl using the Ksp expression:
Ksp = [Ag+][Cl-] = (6.5 * 10^-3) * x
Rearrange the equation and solve for x:
x = Ksp / (6.5 * 10^-3)
Plugging in the values:
x = (1.8 * 10^-10) / (6.5 * 10^-3)
3. Convert the molar solubility to grams per liter (g/L):
The molar solubility (x) you obtained is in moles per liter. To convert it to grams per liter, multiply it by the molar mass of AgCl:
Solubility (g/L) = x * molar mass of AgCl
Plugging in the values:
Solubility (g/L) = (2.77 * 10^-8) * 143.32
So, the solubility of AgCl in a 6.5 * 10^-3 M silver nitrate solution is approximately 3.97 * 10^-6 g/L.