Buffer problems can be quite challenging for many students. It is the purpose of this handout to help guide you through simple buffer problems you might encounter in this course. The best way to solve these problems, and many other problems for that matter, is to follow a step-by-step pattern. I will show you the pattern needed for buffer problems in this document.
Here few questions and answer on acid base buffer calculation with explanation
1. Find the pH of 1.25 M acetic acid and 0.75 M potassium acetate.
Acetic acid kA = 1.74 E-5 pKA = 4.76.
This is a genuine buffer problem. Added to the water are a weak acid and a salt containing the anion of the acid.
There are two good ways to work buffer problems, with the Henderson- Hasselbach equation or with the ionization equilibrium expression of the weak acid or base. I personally have a mental block against the H-H equation because I can never remember whether it uses a positive or negative log and which concentration goes on top. You can use it if you wish. Particularly if you need to calculate buffers often, you should engrave it upon your gray matter. If you really need it and can’t remember it, you can derive it from the ionization equlilbrium expression.[wp_campaign_1]
There are three cautions you need to observe with either equation: (1) Make sure you are using the correct concentration for each variable, (2) check to see if the numbers you propose to use are going to be within the 5% rule for simplification, and (3) estimate the answer from what you know and make sure your final answer is reasonable.
Before actually doing the problem, estimate the answer from your own reasoning. In this case, the pkA of acetic acid is 4.76. The rule is that an equimolar buffer has a pH equal to the pkA and in this problem there is less potassium acetate than acetic acid, so the pH must be lower (more acid) than the pkA within a pH unit or so. If the acetic acid were the only solute, the pH estimate would be the square root of (acid concentration times kA).
The answer should be somewhere between pH of 2.3 and 4.8
The majority of the acetate ion will be from the potassium acetate. Is it right that the total acetate ion concentration will be equal to the concentration of the potassium acetate? Or will the acetate ion concentration from the ionization of the acetic acid contribute more than 5%? The potassium acetate concentration is 0.75 M. The acetate ion concentration from acetic acid would be 0.00466 M, less than 5% of 0.75 M even without the common ion effect. We can safely use 0.75 as the concentration of acetate ion.
Will the concentration of unionized acid be a problem? The measured concentration is 1.25 M and the ionized amount is 0.00466 M, far less than 5% of 1.25 M.
As threatened, we can use the ionization equilibrium expression of acetic acid for the main equation for this problem, substituting for the kA, substituting the concentration of potassium acetate for the concentration of acetate, substituting the concentration of acetic acid, and solving for the hydrogen ion concentration to get the pH.
The answer of pH = 4.5 is a reasonable one by our estimation because it is more acid than the pkA of 4.76.
It is a little easier to do this problem by the Henderson- Hasselbach equation, if you are sure you know it. You must still make sure you are substituting correctly and that your assumptions for simplification are valid (within 5%). The H-H equation is not much good for solutions in which either the acid or ion concentrations are more than ten times one another or in which the concentration of either material is less than one hundred times the kA because it doesn’t easily adapt to a quadratic form.
2. Find the pH of 0.788 M lactic acid and 1.27 M calcium lactate.
Lactic acid kA = 8.32 E-4 pKA = 3.08.
Here is another acid – conjugate base buffer pair. This time there is more conjugate anion than acid concentration, so we expect the pH to be somewhat higher (more alkali) than the pKA. As in the previous problem, there seems to be no complication with either of the components being of too small a concentration or the concentrations being too close to the KA, so there should be no need for a quadratic equation.
The concentration of calcium lactate needs to be doubled (!) to represent the lactate ion concentration because the calcium is divalent and has two lactate ions per formula of calcium lactate. The concenration of acid is more than 100 times the KA, so the concentration of acid is close enough to the concentration of unionized species.
By the ionization equilibrium equation:
Or by the H-H equation, you get the same answer.
3.Find the pH of 0.590 M ammonium hydroxide and 1.57 M ammonium chloride.
ammonium hydroxide kB = 1.78 E-5 pKB = 4.75.
Here we have a weak base and its conjugate cation. We can use the ionization equilibrium expression, but it is different from the acid ionization expression. The ammonium hydroxide ionizes into hydroxide ion and ammonium ion, so it would be best to find the concentration of the hydroxide ion.
The ionization equilibrium expression must have the kB rather than a kA, or the Henderson- Hasselbach equation has to have all its components adapted to alkali, but it is completely analagous to the acid calculation. In either way of doing the problem, you will have to change the answer to the pH
Will we be able to use our standard shortcuts? The concentration of base is more than 100 times the kA, so the measured amount of ammonium hydroxide in solution is a good enough number for the concentration of unionized species. The concentration of weak base and conjugate ion will be within 1:10 of each other, so the amount of conjugate ion can be adequately estimated by the concentration of ammonium chloride. There is high enough concentration of the base so that the ionization of water does not significantly change the hydroxide concentration.
Or by the H-H equation, you get the same answer.
Does the answer make sense? The combination is a base buffer and the pH is slightly base. There is almost three times the concentration of ammonium chloride than ammonium hydroxide, so the pH of the mixture is more acidic than it would be if the buffer had been equimolar. (pH = 9.25)[wp_campaign_2]
4. Explain how to make 5 L of 0.15 M acetic acid-sodium acetate buffer at pH 5.00 if you have 1.00 Molar acetic acid and crystaline sodium acetate.
Here is a problem you may have to actually use one day. In biochemistry some enzymes need to be at a particular pH to work at maximum. You would choose a weak acid with a pkA close to the pH you need. (The pkA of acetic acid is 4.76.) The osmolarity (the total molar amount of dissolved materials) may be specified. (It is here. The total of acetic acid and sodium acetate should be at 0.15 Molar.)
It is most convenient to use the Henderson – Hasselbach equation for this, as it has a term that can be the ratio of the two materials. The form of the H-H equation does not matter, but the concentration of the conjugate ion will have to be greater than the concentration of the acid because the pH is greater than the pkA of the weak acid.
What we get from the H-H equation is the ratio of the two constituents. We can use that ratio as one of the equations in a two – equation – two – unknown setup to substitute one into the other and calculate the concentration of acetic acid, [HA], and the concentration of sodium acetate, [A_].
But we still have not answered the question, “Explain how to make 5 L of pH 5, 0.15 M acetic acid-sodium acetate buffer.” We have a 1.00 Molar solution of acetic acid and crystals of (solid) sodium acetate. The way we have to measure the acetic acid is by measuring the volume of the more concentrated solution. The way to measure the sodium acetate is to weigh it. We would need (54.7885 x 5 = 273.9425) ml of acetic acid and (82.04 x 0.0952115 x 5 = 39.055757) grams of sodium acetate.
The real answer is that you need to weigh 39.1 g of sodium acetate, measure 274 ml of the 1.00 Molar acetic acid and put them into a 5 liter volumetric flask with enough water to dissolve the sodium acetate. Then fill the volumetric flast to the line with distilled water and mix the solution.