Often, stock solutions must be diluted to a desired concentration for a given experiment. Calculating the amount of buffer solution you need to add for the dilution is fairly straightforward but can take some practice to learn. You may also be required to calculate the relative proportions of each species in a buffered solution, a calculation that involves the Henderson-Hasselbach equation. To solve these kinds of problems, follow the steps outlined below.
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Determine what units the concentration of each solution is expressed in. Concentrations are usually written using either molarities or % concentration, where molarity denotes moles of solvent per litre of solution and % denotes grams of solute per 100 millilitres of solution. It doesn't matter which unit you use, but make sure that all figures in your calculation use the same units. If you try to use molarity for one species and % for the other, your answer will be wrong.
Solve dilution problems using the simple equation C1 V1 = C2 V2, where C1 is the starting concentration, V1 is the initial volume of the solution, C2 is the final concentration and V2 is the final volume. If you wanted to use 100 millilitres of 1 molar solution to make a 0.1 molar solution, for example, you would plug 100ml in as V1, 1 M as C1 and 0.1 as C2, then solve for V2. By subtracting V1 from V2 you can then determine how much buffer you need to add for your dilution.
Solve buffer capacity problems using the Henderson-Hasselbach equation, pH = pKa + log ( [A-] / [HA] ). In this equation, the pKa is an experimentally determined value that denotes the strength of an acid, while [HA] is the concentration of the acid and [A-] is the concentration of the conjugate base of this acid. If you know the pH of your buffered solution and the pKa of your buffer compound, you can solve for the ratio [A-] / [HA], then multiply this figure by the total amount of buffer compound in your solution to determine how much is present in conjugate base form. Subtract this value from the total amount present and you now have the amount present in protonated form as well.
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