The molarity of a solution is equal to moles of solute per liter of solution. In other words:
\(M = \frac{n}{V}\)
Where M is molarity, n is moles of solute, and V is the volume of soultion in liters. For example, what is the molarity of 50 g NaCl placed in 2 L of water? First, we calculate the moles of solute:
\(\ce{50 g NaCl * (1 mol / 58.44 g) = 0.86 mol NaCl}\)
Next, we divide n moles of solute by the total volume of solution to get the molarity.
\(M = \frac{n}{V} = \frac{0.86 \, \text{mol}}{2 \, \text{L}} = 0.43 \, \text{M}\)
Molarity can be expressed using metric SI prefixes similar to other units (e.g. one can express distance as meters (m) of kilometers (km)). For example, consider the following: What is the molarity of 0.1 g of NaCl in 1 L of solution?
\(M = \frac{0.0017 \, \text{mol}}{2 \, \text{L}} = 0.00085 \, \text{M}\, = 0.85 \, \text{mM}\)
Here, since we have such a small amount of NaCl present, the molarity is much lower than before. It is easier to express this with units of millimolar (mM), or 1/1000th (10-3) moles per L. Multiplying the above by 1,000 (there are 1,000 millimoles in 1 mole) gives us a 0.85 mM solution of NaCl, which is now expressed in units that are more appropriate for the smaller concentration.
What happens if we take some amount of a solution on known concentration and place it into a larger total volume? We dilute our original solution, making the overall concentration less. Such a process is necessary in many chemical and biological research applications. A stock solution of known concentration can be diluted to make additional solutions at the desired concentration. Consider one of the earlier solutions we looked at: 0.86 M NaCl. What would happen if you took 100 mL of this solution and put it into a new vessel containing 500 mL of water? First, let's determine how many moles of NaCl we would be transferring:
\(0.1 L \times 0.86 M = 0.086 \text{ moles}\)
Since molarity is moles/liter, multiplying the molarity by the volume gives us the total number of moles. From here, we divide this numbers of moles (0.086) by our new total volume of 600 mL (100 mL of 0.86 M solution + 500 mL water) to get the resulting molarity:
\(\frac{0.086 \text{ moles}}{0.6 L} = 0.143 M\)
The Ksp, or solubility product constant, represents the equilibrium contant for a solid in an aqueous solution. The greater the solubility, the higher the Ksp. To calculate the Ksp, we take the molarities of the products and multiply them, while raising the product to its coefficient power.
\(A(s) \rightleftharpoons bB(aq) + dD(aq)\)
\(K_{sp} = [B]^b[D]^d\)
Note that solids are not included here. Solids are not included in equilibrium constant expressions as changes in their concentrations are insignificant. Consider the following problem:
For the dissolution of silver chloride (AgCl) in water:
\(AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)\)
If the solubility of AgCl is 1.34 x 10-5 M at 25°C, calculate the Ksp AgCl.
\(K_{sp} = [Ag^+][Cl^-] = (1.34 \times 10^{-5})^2 = 1.80 \times 10^{-10}\)