how many grams of CaCO3 will dissovle in 300ml of 0.05M Ca(NO3)2 ?
Ksp of CaCO3 : 4.7x10^-9
M.W of CaCO3 100.01g/mol
To determine the number of grams of CaCO3 that will dissolve in 300 mL of 0.05M Ca(NO3)2, we can use the solubility product constant (Ksp) of CaCO3.
First, let's write the balanced chemical equation for the dissociation of CaCO3:
CaCO3 ↔ Ca2+ + CO3^2-
The Ksp expression for CaCO3 is as follows:
Ksp = [Ca2+][CO3^2-]
Given that the molarity of Ca(NO3)2 is 0.05M and assuming complete dissociation, the concentration of Ca2+ is also 0.05M.
Now, let's assume that 'x' moles of CaCO3 dissolve, which means that 'x' moles of Ca2+ and 'x' moles of CO3^2- are formed.
Using stoichiometry, we can calculate the concentration of CO3^2-:
Since one molecule of CaCO3 forms one CO3^2- ion:
[CO3^2-] = x (moles/L)
Using the volume of the solution (300 mL = 0.3L) and the concentration calculated, we can determine the moles of CaCO3 that will dissolve:
moles of CaCO3 = [CO3^2-] = x (moles/L)
moles of CaCO3 = [CO3^2-] × volume = x × 0.3L
Finally, using the molar mass of CaCO3 (100.01 g/mol), we can calculate the grams of CaCO3 that will dissolve:
grams of CaCO3 = moles of CaCO3 × molar mass = (x × 0.3L) × 100.01 g/mol
To find 'x', we need to rearrange the Ksp expression as follows:
Ksp = [Ca2+][CO3^2-]
4.7x10^-9 = 0.05M × x (x represents the concentration of CO3^2- in moles per liter)
Now, solve the equation for x:
x = (4.7x10^-9) / 0.05M
Calculate 'x' using the above equation, and substitute it into the formula to find the grams of CaCO3 that will dissolve.
To find out how many grams of CaCO3 will dissolve in 300 ml of 0.05 M Ca(NO3)2, we need to calculate the solubility product (Ksp) of CaCO3 and compare it to the concentration of Ca(NO3)2.
First, let's calculate the number of moles of Ca(NO3)2 in 300 ml of 0.05 M solution:
Molarity (M) = Moles (mol) / Volume (L)
0.05 M = Moles / 0.3 L
Rearranging the equation, we find:
Moles (mol) = 0.05 M x 0.3 L
Moles (mol) = 0.015 mol
Since Ca(NO3)2 dissociates into two moles of Ca2+ ions for every mole of Ca(NO3)2, the number of moles of Ca2+ ions is 0.015 mol x 2 = 0.03 mol.
Next, we need to calculate the maximum number of moles of CaCO3 that could be formed from the Ca2+ ions present. Since CaCO3 dissociates into one mole of Ca2+ ion and one mole of CO3^2- ion, the maximum number of moles of CaCO3 is also 0.03 mol.
Finally, we can calculate the mass of CaCO3 using its molar mass:
Mass (g) = Moles (mol) x Molar mass (g/mol)
Mass (g) = 0.03 mol x 100.01 g/mol
Mass (g) = 3.0003 g
Therefore, approximately 3.0003 grams of CaCO3 will dissolve in 300 ml of 0.05 M Ca(NO3)2 solution.