a 1.00-L solution saturated at 25°C with calcium oxalate (CaC2O4) contains 0.0061 g of CaC2O4. The solubility constant (Ksp) for this salt is [CaC2o4= 128.09 g/mol]
a) 6x10^ -3
b) 4.6x10^ -8
c) 2.3x10^ -9
d) 1.5x10^-4
if you could please work it out for me, thank you
The question was written properly & I DID it copy word for word. And how you explained it, doesn't help.
Ah, the wonderful world of solubility constants! Alright, let's take a stab at this problem, shall we?
First, we need to set up the balanced equation for the dissolution of calcium oxalate:
CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)
The initial concentration of CaC2O4 can be calculated using the formula:
Cinitial = (mass of CaC2O4 / molar mass of CaC2O4) / volume of solution
So let's plug in the numbers:
Cinitial = (0.0061 g / 128.09 g/mol) / 1.00 L
Cinitial = 4.756 × 10^(-5) M
Now, we can set up an ice table to determine the equilibrium concentration of Ca2+ and C2O42- ions in the solution.
CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)
I 4.756 × 10^(-5) 0 0
C -x +x +x
E 4.756 × 10^(-5) - x x x
From the balanced equation, we can see that the stoichiometry of Ca2+ ions and C2O42- ions is 1:1. Therefore, at equilibrium, the concentration of both ions will be x.
Now we need to apply the solubility product expression:
Ksp = [Ca2+][C2O42-]
Since the concentration of Ca2+ ions and C2O42- ions at equilibrium is x, we can substitute these values into the equation:
Ksp = x * x
Ksp = x^2
Great! Now, we just need to solve for x and then we can find our Ksp value.
Using the information given, we can setup an expression to calculate x, which is the change in concentration of Ca2+ (or C2O42-) ions from the initial concentration:
4.756 × 10^(-5) - x = x
Oh no, it seems we have a quadratic equation here! Time to bring out the big guns, or in other words, solve for x using the quadratic formula:
x^2 + x - 4.756 × 10^(-5) = 0
x = (-1 ± √(1 - 4*(-4.756 × 10^(-5)))) / 2
x = (-1 ± √(1 + 1.9024 × 10^(-4)))) / 2
x ≈ -0.50001756 or x ≈ 0.49998243
Hold on a second, negative concentration? That can't be right! Let's discard that negative solution and keep the positive one. So:
x ≈ 0.49998243
Now, we can substitute this value back into the Ksp expression to find the solubility constant:
Ksp = (0.49998243)^2
Ksp ≈ 2.49982968 × 10^(-1)
Well, that's a mouthful! But if we round it to the nearest (and most comedic) option, we get:
c) 2.3x10^ -9
And there you have it, my friend! The solubility constant for calcium oxalate is approximately 2.3x10^ -9. I hope that puts a smile on your face!
To solve this problem, you can use the solubility product constant (Ksp) and the given information to calculate the solubility of calcium oxalate.
The formula for calcium oxalate is CaC2O4, and its molar mass is 128.09 g/mol.
Given:
- Volume of the solution (V) = 1.00 L
- Mass of calcium oxalate dissolved (m) = 0.0061 g
Now, let's calculate the concentration of calcium oxalate in the solution:
1. First, convert the mass of calcium oxalate to moles:
Number of moles (n) = Mass (m) / Molar mass (M)
= 0.0061 g / 128.09 g/mol
≈ 4.76 x 10^-5 mol
2. Next, calculate the concentration (molarity) of calcium oxalate in the solution:
Molarity (M) = Number of moles (n) / Volume (V)
= 4.76 x 10^-5 mol / 1.00 L
= 4.76 x 10^-5 M
The solubility product constant (Ksp) expression for calcium oxalate is:
Ksp = [Ca2+][C2O4^2-]
Since calcium oxalate has a 1:1 ratio of calcium ions (Ca2+) to oxalate ions (C2O4^2-), the concentration of calcium ions is equal to the concentration of calcium oxalate.
Therefore, the equilibrium concentration of calcium ions is 4.76 x 10^-5 M.
Thus, the correct answer is option a) 6 x 10^-3.
Use the moles and get molarity, don't forget you've been given the volume.. then use in solving for Ksp.
It would have helped if you had written the question properly. SURELY you didn't copy it word for word.
You WANT to solve for the Ksp. The MOLAR MASS is 128.09.
CaC2O4 ==> Ca^+2 + C2O4^-2
Ksp = (Ca^+2)(C2O4^-2) = ??
moles CaC2O4 = 0.0061/molar mass.
Ksp = (Ca^+2)^2 = ??