Concentration Calculations

More on "ppm" ...

Remember that ppm is parts-per-million and is like percentage except that it is based on 1 million parts (not 100). So by mass,

\[\rm ppm = {mass\;of\;solute\over mass\;of\;solution} \times 10^6 \]

In dilute aqueous solutions, the density is the same as water (1 g/mL) which means that \({\rm 1\;ppm \approx 1\; mg/L}\). So anytime you see "ppm" change it to "mg/L" for all our calculations. Then you can easily convert that to mol/L to get concentration in molarity.

For the rest of this page, concentration is assumed to be in molarity although most all concentration terms will work out the same way in that they are all intensive properties.

Molarity

• Concentration in molarity (mol/L or M) is often depicted with square brackets around the solute of interest. For example, the concentration of hydrogen ion is depicted as [H+].
• Concentration is also sometimes shown with a capital "\(C\)" or "\(M\)" and a subscript for the solute. Using this designation, the concentration of hydroxide would be \(C_{\rm OH-}\) or \(M_{\rm OH-}\) .

Dilutions - adding more solvent (ie. water) to an already existing solution

\[M_1\cdot V_1 = M_2\cdot V_2\]

ANY concentration and volume unit will work here as long as the same units are on BOTH sides of the equation.

Titrations (1:1 ratios)

For simple 1:1 ratios of acid and base: \({\rm A}+{\rm B}\rightarrow {\rm water\;+\;salt}\)

\[M_{\rm acid}\cdot V_{\rm acid} = M_{\rm base}\cdot V_{\rm base}\]

NOTE: This equation only works for monoprotic acids (one proton in the formula) and monobases (one hydroxide in the formula). Why? Because then the moles of acid will equal the moles of base, a nice 1:1 ratio. If you have ANY other ratio (not 1:1) you must use the following formula:

Titrations (any ratio)

For any acid/base ratio (\(a:b\)) of acid and base: \(a{\rm A}+b{\rm B}\rightarrow {\rm water\;+\;salt}\)

\[{M_{\rm acid}\cdot V_{\rm acid}\over a} = {M_{\rm base}\cdot V_{\rm base}\over b}\]

Note that \(a\) and \(b\) are the coefficients for the acid and base in the balanced chemical equation.

More Volumetric Stuff - getting moles or millimoles

For a given volume of any solution, you can always calculate the number of moles of solute...

\[M_{\rm solute}\cdot V_{\rm solution} = {\rm moles\;of\;solute} = n_{\rm solute}\]

To get moles you must use molarity for the concentration and liters as the volume. If you use milliliters (mL) for volume you will get millimoles (mmol) for the amount. Be careful and note which type unit you are using. Dr. McCord will often use mL with molarity and get mmol for \(n\). Always make sure you know what units you are using.

If you pour together multiple solutions of the same solute, you must calculate the number of moles in each separate solution first, then add all the moles together, and divide by the total combined volume. As long as you quantify all the mole sources (whether solutions, solids, or whatever) and the total accumulated volume, you can easily calculate the concentration.

Mixing Together 3 solutions of the same solute

Mix solutions A, B, and C (all different concentrations and volumes, but all the same solutes)

\[M_{\rm mixture} = {M_{\rm A}V_{\rm A} + M_{\rm B}V_{\rm B} + M_{\rm C}V_{\rm C} \over V_{\rm A}+V_{\rm B}+V_{\rm C}}\]

Note that the numerator here is just a total (sum) of all the moles of solute and the denominator is a total (sum) of all the volumes.

Acid/Base Formulas

Be sure and know how to get any one of these values from any one of the others...

\({\rm pH} = -\log[{\rm H}^+]\)		\({\rm pOH} = -\log[{\rm OH}^-]\)
\([{\rm H}^+] = 10^{\rm -pH}\)		\([{\rm OH}^-] = 10^{\rm -pOH}\)
\(K_{\rm w} = [{\rm H}^+][{\rm OH}]^-\)		\({\rm pH+pOH = 14}\)
\(K_{\rm w} = 1.0\times 10^{-14}\)

the value for \(K_{\rm w}\) is only valid at 25 C