The Buffer equation

Knowing an acids strength exponent pKa it is also possible to calculate pH in a buffer solution and this is the subject for the following section. We take a starting point in the well-known acid-base reaction  

The equilibrium expression for the general acid-base reaction may be expressed as  

The equation for Ka is rewritten in order to be able to contain pH. This happens by applying the decimal logarithm on both sides of the equilibrium sign and do a few simple manipulations. Hereby the buffer equation arises  

The buffer equation is also known as the Henderson-Hasselbalch equation. This equation is in principle just another version of the expression for Ka but it may nevertheless other be easier to apply. Using the buffer equation one must remember that HA and A denotes the corresponding acid-base pair and that pKa is the acid exponent of the acid (HA). In the following example the buffer effect is illustrated in a buffer system consisting of equal amounts of acetic acid and acetate into which strong base is added. 

A buffer solution consists of 0.5 M acetic acid CH3COOH (Ka value of 1.8 10 M) and 0.5 M sodium acetate CHsCOONa. This solution consists of a weak acid (acetic acid) ad its corresponding weak base (acetate ion). As the amount of the weak acid and weak base are similar (it is sufficient if there just in the same order of magnitude) we have a buffer system. 

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The buffer equation, which is often called the Henderson-Hasselbalch equation, is used to calculate the equilibrium pH of a buffer solution directly from initial concentrations. The approximation is valid as long as the difference between initial concentrations and equilibrium concentrations is negligibly small. As a rule of thumb, the buffer equation can be applied when initial concentrations of H j4 and A differ by less than a factor of 10. Example provides an illustration of the use of the buffer equation. 

We use the seven-step strategy for equilibrium problems, except that we identity this as a buffer solution. This allows us to use the buffer equation in place of an equilibrium constant expression. 

This proton transfer reaction involves the second acidic hydrogen atom of carbonic acid, so the appropriate equilibrium constant is. a 2 > whose p is found in Appendix E p. a 2 — 10.33. Because this is a buffer solution, we apply the buffer equation ... 

Now substitute the appropriate values into the buffer equation and evaluate ... 

The first four steps of the seven-step strategy are identical to the ones in Example. In this example, addition of a strong acid or base modifies the concentrations that go into the buffer equation. We need to determine the new concentrations (Step 5) and then apply the buffer equation (Step 6). In dealing with changes in amounts of acid and base, it is often convenient to work with moles rather than molarities. The units cancel in the concentration term of the buffer equation, so the ratio of concentrations can be... 

Einally, substitute these new amounts into the buffer equation to compute the new pH ... 

Use the seven-step strategy to calculate the pH of the buffer solution using the buffer equation. Then compare the amount of acid in the solution with the amount of added base. Buffer action is destroyed if the amount of added base is sufficient to react with all the acid.The buffering action of this solution is created by the weak acid H2 PO4 and its conjugate base HP04. The equilibrium constant for this... 

Substitute the tabulated value of p 2 and the given concentrations into the buffer equation to... 

The buffer equation indicates that the pH of a buffer solution is close to the p of the acid used to prepare the... 

Because we know we are dealing with a buffer solution made from a specific conjugate acid-base pair, we can work directly with the buffer equation. We need to calculate the ratio of concentrations of conjugate base and acid that will produce a buffer solution of the desired pH. Then we use mole-mass-volume relationships to translate the ratio into actual quantities. 

What concentrations of NH4 and NH3 are required First use the buffer equation to find the proper ratio of base to acid, then use the total molarity of the solution to determine the concentrations of NH3... 

During most of the titration, both H A and its conjugate base. A, are major species. As a result, the solution is buffered, and the pH changes relatively slowly with added hydroxide. In this region, the pH of the solution can be calculated using the buffer equation. 

This buffer region contains the midpoint of the titration, the point at which the amount of added OH" is equal to exactly half the weak acid originally present. In the current example, the solution at the midpoint contains 0.0375 mol each of acetic acid and acetate. Applying the buffer equation reveals the key feature of the midpoint ... 

The titration curve does not give directly. However, at the midpoint of the titration the concentration of Ep and EpH+ are identical. Use this information in the buffer equation to show that at the midpoint of the titration, the pH of the solution equals the p for EpH ... 

Within the first buffer region, both H2 A and RA are major species in solution, and we can apply the buffer equation to caicuiate the pH. Haifway to the first equivalence point of the titration [H2 A] — [H A ] and pH pTai = 1.82. 

Moving beyond the first stoichiometric point, the titration enters the second buffer region. Here, the major species are H A and its conjugate base,. The pH in this region is given by the buffer equation, using the p of H j4.". At the second midpoint, [H j4" ] = [j4 " ], and pH = p. Ta 2 For the titration of maleic acid, pH = 6.59 at the second midpoint. 

When plotted on a graph of pH vs. volume of NaOH solution, these six points reveal the gross features of the titration curve. Adding additional calculated points helps define the pH curve. On the curve shown here, the red points A-D were calculated using the buffer equation with base/acid ratios of 1/3 and 3/1. Point E was generated from excess hydroxide ion concentration, 2.00 mL beyond the second stoichiometric point. You should verify these additional five calculations. 

K0bs =1.3 when Mg2+ was omitted from the buffer. Equation (21.35) allows 9 > 0 when [Mg2 = 0 when this equation was used to fit the simulated data in Fig. 21.5, much better fits to the data were obtained than with Eq. (21.34) (see residuals plotted in the lower panels of Fig. 21.5), and the Ar2+ value calculated from the fitted parameters reproduced the values used to generate the curves. 

This is the Henderson-Hasselbach equation, or simply the buffer equation. We can use the buffer equation to calculate the pH of a buffer. We can also use it to determine the ratio of weak acid to conjugate base at a given pH. 

Let s try another one. What is the pH of a buffer containing 0.25 M ammonia and 0.75 M ammonium chloride Well, the weak acid in this case is the ammonium ion. The chloride ion is a spectator to be ignored. Ammonia is a weak base, and the conjugate base of the ammonium ion. So, since this solution contains a weak conjugate acid-base pair, it is a buffer, and we can calculate the pH using the Henderson-Hasselbach equation. The Henderson-Hasselbach equation calls for the pKa of the acid so in this case, we need the pfor the ammonium ion. The pKh for ammonia is 4.74, so the pKa for the ammonium ion is 9.26. If we substitute this value and the values for the concentrations into the buffer equation, we find the pH of this solution is 8.78. 

One last example, and this one is special. What is the pH of a buffer that contains acetic acid and sodium acetate both at a concentration of 0.30 M When we put the numbers into the buffer equation, we see the ratio of acetate ion to acetic acid is 1. The logarithm of 1 is 0, so the pH is equal to the pkO This is true of any buffer system. When the concentrations of the weak acid and its conjugate base are equal, the pH equals the pKa. 

A quick look at the buffer equations explains this. The pH changes from the pKa by the log of the ratio of the weak base to the weak acid. If the pH is one unit above the pFCa, the ratio of the weak base to weak acid is 10. This is still an effective buffer, because there is still an appreciable concentration of both the weak acid and the weak base. 

The pK3 of dihydrogenphosphate is the closest to our desired pH. So, the weak conjugate acid-base pair we need is dihydrogenphosphate and monohydrogenphosphate. If we substitute in the pH and pKa values into the buffer equation, we have ... 

The shape of the curves in an alpha plot is determined by the buffer equation. At low pH, most of the material is present as its conjugate acid. As the pH increases, the conjugate acid form converts to the conjugate base form. At high pH, most of the material is present in the conjugate base form. 

June 3, 1978, Lynn, Massachusetts, USA - Feb. 10, 1942 Boston, USA) Henderson studied medicine at Harvard and was Professor of Biological Chemistry at Harvard University, Cambridge, Massachusetts, from 1904 to 1942 [i]. Henderson published on the physiological role of -> buffers [ii-vii] and the relation of medicine to fundamental science. Because he and also - Hasselbalch made use of the law of mass action to calculate the - pH of solutions containing corresponding acid-base pairs, the buffer equation is frequently (esp. in the biological sciences) referred to as -> Henderson-Hasselbalch equation. 

As bisulfite ion is used up in reaction (7), the hydrogen-ion concentration adjusts itself according to the buffer equation (8). If an indicator such as phenolphthalein is added to the mixture, it will undergo a sudden color change when the pH of the solution reaches the pA, of the indicator. The time t required for the color change is related to the rate constant for reaction (7) by the equation ... 

The logarithm is in numerical-base-10. This is also called the buffer equation. 

Many assumptions are required to use these equations, but other methods of calculating the pH of buffers are too difficult to appear on an examination and are beyond the scope of high school chemistry. Therefore, if you are asked for the pH of a buffer solution on the teacher certification exam, you can use the buffer equation with confidence. 

The initial concentrations of acetic acid and the acetate ion are equally large which is why pH in this case is simply equal to the pKa value of acetic acid according to the buffer equation ... 

This pH value could also be found with the use of the buffer equation as we still have a buffer system with a weak corresponding acid-base pair of the same order (x is veiy small) ... 

Acetic acid has a pKa value of 4.74. In this case [CH3COOH] = [CH3COO ] before HCl is added which is why pH initially may be found via the buffer equation ... 

We still have a buffer system as the corresponding weak acid-base pair is present in amounts of the same magnitude. We may therefore calculate pH using the buffer equation ... 

On the way to the first point of equivalence we have a buffer system consisting of the weak acid H2A and the corresponding weak base HA" which is why pH does not change very much. Halfway towards the first point of equivalence pH I calculated form the buffer equation ... 

Further we introduced buffer chemistry and saw how pH may be calculated in buffer solutions and on how the buffer equation is often used in practice. When one has a solution of a weak acid and its corresponding weak base, both in concentrations of the same magnitude, one has a buffer system and the buffer equation may be used to calculate pH. Lastly, we looked at titration and on pH curves exemplifyed through examples of titration of monovalent weak acid with strong base and titration of divalent acid likewise with strong base NaOH. In the end we saw how colour indicator work. 

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