A mixture is prepared by adding 0.25 L of 2.5 x 10^-3 M MgCl2 (aq) to 0.20 L of 0.010 M H2CO3 (aq). Does a precipitate form when the two solutions mix?
Determine the molarity of a solution prepared by dissolving 16.9g of naoh in enough water to make 250.0ml of solution
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To determine if a precipitate forms when the two solutions mix, we need to compare the solubility products (Ksp) of the ions involved.
The solubility product for magnesium carbonate (MgCO3) can be calculated using the equation:
Ksp = [Mg2+][CO3^2-]
To find the concentration of Mg2+ in the mixture, we first need to calculate the number of moles of MgCl2. Given that the volume is 0.25 L and the concentration is 2.5 x 10^-3 M, we can use the formula:
moles = volume x concentration
moles = 0.25 L x 2.5 x 10^-3 M
moles = 6.25 x 10^-4 mol
Since MgCl2 is a 1:1 electrolyte, the concentration of Mg2+ ions in the solution is equal to the number of moles of MgCl2:
[Mg2+] = 6.25 x 10^-4 mol / (0.25 L + 0.20 L)
[Mg2+] = 1.25 x 10^-3 M
Now, let's calculate the concentration of CO3^2- ions in the solution. Since H2CO3 is a weak acid that dissociates into H+ and HCO3-, the concentration of CO3^2- ions can be calculated using the equation:
[CO3^2-] = [HCO3-]
[HCO3-] is the same as the original concentration of H2CO3, which is 0.010 M.
Now, let's compare the solubility product (Ksp) to determine if a precipitate forms. The Ksp for MgCO3 is given as 5.6 x 10^-7.
Ksp = [Mg2+][CO3^2-]
Ksp = (1.25 x 10^-3 M)(0.010 M)
Ksp = 1.25 x 10^-5
Since the calculated Ksp (1.25 x 10^-5) is greater than the given Ksp (5.6 x 10^-7), a precipitate will form when the two solutions mix.