OOPs i Meant How many grams CaCo3will dissolve on 3.0*10^2mL of 0.050 M Ca(NO3)2? Ksp=8.7*10^-9
Well, well, looks like you're in need of some chemistry help! Don't worry, Clown Bot is here to bring some laughter into your equation.
To solve this problem, we need to use the concept of solubility product constant (Ksp). But before we dive into those boring calculations, let's try to lighten up the mood, shall we?
Why did the chemist dissolve calcium carbonate in water? Because it’s nice to have a little "lime" with your "H2O"!
Okay, now let's get serious. The balanced equation for the dissolution of calcium carbonate (CaCO3) in water is:
CaCO3 (s) ⇌ Ca2+ (aq) + CO3^2- (aq)
The equilibrium expression for this reaction is:
Ksp = [Ca2+] [CO3^2-]
So, to find out how many grams of CaCO3 will dissolve, we need to calculate the concentration of Ca2+ and CO3^2- ions.
Given that the volume of Ca(NO3)2 solution is 3.0 * 10^2 mL or 300 mL and its concentration is 0.050 M, we can set up the following equation:
0.050 M = [Ca2+]
Since Ca(NO3)2 dissociates into one Ca2+ ion and two NO3- ions, the concentration of Ca2+ ions is equal to the initial concentration of Ca(NO3)2.
Now, let's use the Ksp expression to find the concentration of CO3^2- ions:
Ksp = [Ca2+] [CO3^2-]
8.7 * 10^-9 = (0.050 M) [CO3^2-]
Solving for [CO3^2-]:
[CO3^2-] = (8.7 * 10^-9) / (0.050 M)
Now that we have the concentration of CO3^2- ions, we can calculate the mass of CaCO3 that will dissolve using its molar mass.
But hey, before we wrap things up, let me share another chemistry joke with you:
Why did the chemist like studying carbonates? Because they were always full of fizz-cal excitement!
Okay, getting serious again. The molar mass of CaCO3 is 100.0869 g/mol.
Now we have all the pieces of the puzzle to find the mass of CaCO3 that will dissolve:
Mass = molar mass * moles
The moles can be calculated from the concentration using the equation:
moles = concentration * volume (in liters)
You can plug in the values and calculate the final answer.
And there you have it! I hope my humor didn't dissolve your concentration while solving the problem. Remember, laughter is the best catalyst when it comes to chemistry!
To calculate the number of grams of CaCO3 that will dissolve in 3.0 * 10^2 mL of 0.050 M Ca(NO3)2, we need to use the solubility product constant (Ksp) for CaCO3.
The balanced equation for the dissociation of CaCO3 is:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
The molar solubility of CaCO3 in terms of x is:
CaCO3(s) ⇌ xCa2+(aq) + xCO3^2-(aq)
According to the equation, 1 mole of CaCO3 will produce 1 mole of Ca2+ and 1 mole of CO3^2-.
The concentration of Ca2+ in the solution is equal to the concentration of Ca(NO3)2, which is 0.050 M.
Therefore, the concentration of Ca2+ in terms of x is:
[Ca2+] = x
Since 1 mole of CaCO3 will produce 1 mole of Ca2+, the concentration of Ca2+ is also equal to x.
The concentration of CO3^2- in the solution is also equal to x.
Using the Ksp expression for CaCO3:
Ksp = [Ca2+][CO3^2-]
Substituting the concentrations:
Ksp = (x)(x) = x^2
Given that the Ksp for CaCO3 is 8.7 * 10^-9, we can set up the equation:
8.7 * 10^-9 = x^2
Taking the square root of both sides, we find that:
x = √(8.7 * 10^-9) ≈ 9.34 * 10^-5 M
Now, we can calculate the moles of CaCO3 that will dissolve in 3.0 * 10^2 mL (0.300 L) of the Ca(NO3)2 solution:
moles of CaCO3 = (9.34 * 10^-5 M)(0.300 L) = 2.80 * 10^-5 mol
Finally, to convert moles to grams, we need to multiply by the molar mass of CaCO3, which is 100.09 g/mol:
grams of CaCO3 = (2.80 * 10^-5 mol)(100.09 g/mol) ≈ 2.800 g
Therefore, approximately 2.800 grams of CaCO3 will dissolve in 3.0 * 10^2 mL of 0.050 M Ca(NO3)2.
To determine how many grams of CaCO3 will dissolve in 3.0*10^2 mL of 0.050 M Ca(NO3)2, we need to use the concept of solubility product constant (Ksp) and the balanced chemical equation for the dissociation of calcium carbonate.
The balanced chemical equation for the dissociation of calcium carbonate (CaCO3) is:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
From the equation, we can see that one CaCO3 molecule dissociates into one calcium ion (Ca2+) and one carbonate ion (CO3^2-).
To solve the problem, we can follow these steps:
1. Calculate the molar concentration of Ca2+ ions in the Ca(NO3)2 solution.
- The molar concentration (C) is given as 0.050 M, which means there are 0.050 moles of Ca2+ ions per liter (L) of the solution.
- Convert the volume of the solution to liters: 3.0*10^2 mL = 3.0*10^2 / 1000 L = 0.3 L.
- Therefore, the moles of Ca2+ ions in the solution = C * V = 0.050 M * 0.3 L = 0.015 moles.
2. Since the dissociation ratio between Ca(NO3)2 and CaCO3 is 1:1 (one mole of CaCO3 produces one mole of Ca2+ ions), the moles of CaCO3 that will dissolve are also 0.015 moles.
3. Calculate the grams of CaCO3 that will dissolve.
- Convert the moles of CaCO3 to grams using its molar mass: 1 mole of CaCO3 = 100.1 g (approximate molar mass of CaCO3).
- Therefore, the grams of CaCO3 that will dissolve = 0.015 moles * 100.1 g/mole ≈ 1.5015 g.
So, approximately 1.5015 grams of CaCO3 will dissolve in 3.0*10^2 mL of 0.050 M Ca(NO3)2 solution.
Note: The given Ksp value (8.7*10^-9) is not directly used in this calculation. It is the solubility product constant, which represents the equilibrium constant for the dissociation of CaCO3 and can be used to calculate the solubility limit of the compound under specific conditions.
CaCO3 ==> Ca^+2 + CO3^-2
S = solubility
..S........S....... S
Ca(NO3)2 ==> Ca^+2 + 2NO3^-
..0.05M......0.05M....0.10M
Ksp = (Ca^+2)(CO3^-2) = (S+0.05)(S) = Ksp.
Solve for S (which will be moles/L), then convert to moles in 300 mL, then convert to grams by g = moles x molar mass.
Post your work if you get stuck.