Calculate the solubility product. The equilibrium is given below:. Notice that each mole of barium sulfate dissolves to give 1 mole of barium ions and 1 mole of sulfate ions in solution. That means that:. Substitute these values into the solubility product expression, and simplify:. The known K sp values from the Table above can be used to calculate the solubility of a given compound by following the steps listed below.
The K sp of calcium carbonate is 4. The variable will be used to represent the molar solubility of CaCO 3. Therefore, the equilibrium concentrations of each ion are equal to. The K sp expression can be written in terms of and then used to solve for. The concentration of each of the ions at equilibrium is 6. We can use the molar mass to convert from molar solubility to solubility. Before a solubility limit can be applied as a conversion factor, each substance that is referenced in the given problem must first be classified as a solute or a solvent.
As stated in Section 7. Since the chemical formula for water, H 2 O, is associated with the In order to determine whether this solution is saturated or unsaturated, the solubility of hydrogen sulfide, H 2 S, which has a reported value of 0. The chemical formula for hydrogen sulfide, H 2 S, must be incorporated into the numerator of this conversion, as shown below, in order to achieve the desired unit transformation.
Finally, because solubility limits correspond to the exact amount of solute that can dissolve in a corresponding amount of solvent, the quantities that are calculated using a solubility proportion should not be rounded. Based on this calculated value, exactly 2. Since the solution that is described in the problem is prepared using 1.
So the equilibrium concentration of calcium two plus ions is zero plus X, or just X, and the equilibrium concentration of fluoride anions will be zero plus 2X, or just 2X. The next step is to write the Ksp expression from the balanced equation. So Ksp is equal to the concentration of calcium two plus ions, and since there's a coefficient of one in the balanced equation, that's the concentration of calcium two plus ions raised to the first power, times the concentration of fluoride anions, and since there is a coefficient of two in the balanced equation, it's the concentration of fluoride anions raised to the second power.
Pure solids are not included in equilibrium constant expression. So we're going to leave calcium fluoride out of the Ksp expression.
The concentration of ions in our Ksp expression are equilibrium concentrations. Therefore we can plug in X for the equilibrium concentration of calcium two plus and 2X for the equilibrium concentration of fluoride anions. We can also plug in the Ksp value for calcium fluoride.
So that would give us 3. Next we need to solve for X. So, 3.
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