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17-2: Controlling pH: Buffer Solutions

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    A buffer is a solution that maintains a constant pH when an external acid or base is added to it. This is done by having an internal acid and base within the buffer solution, but, the internal acid and base can not neutralize each other.

    Three Requirements for a Buffer

    1. Internal Acid- that neutralizes an external base that has been added to the solution, preventing the base's hydroxide from raising the pH.
    2. Internal Base- that neutralizes an external acid that has been added to the solution, preventing acid's hydronium from lowering the pH.
    3. The internal acid and base can't neutralize each other.

    An acid/base conjugate pair fulfills these requirements: Thus there are two types of buffers

    Two Types of Buffers

    Buffer Type I: A Weak acid and its conjugate base

    \[HA \leftrightharpoons H_3O^+ + A^-\]

    Adding a Base converts HA to A-.
    ex: \[HA + NaOH \rightarrow NaA + H_2O\]

    note, we often omit the spectator ions (sodium) in these equations, and are implying that A- is the anion of a salt.


    Adding an Acid converts A- to HA.
    ex:\[HCl + A^- \rightarrow HA + Cl^-\]


    The total concentration of an acid and its conjugate base are constant, it is just shifting between the protonated and nonprotonated form.

    note, we often omit the sodium (spectator ion) in these equations

    Buffer Type II: A Weak base its conjugate acid

    \[B+H_2O \leftrightharpoons BH^+ + OH^-\]

    Adding an Acid converts B to BH+.

    ex: \[HCl + NH_3 \rightarrow NH_4^+ +Cl^-\]


    Adding a Base converts BH+ to B.

    ex:\[HCl + A^- \rightarrow HA + Cl^-\]


    The total concentration of an acid and its conjugate base are constant, it is just shifting between the protonated and nonprotonated form.

    Note, we often omit the sodium (spectator ion) in these equations

    Two Ways to Make a Buffer

    1. Mix a weak acid (or base) with a soluble salt of its conjugate.
      A solution that is made by diluting 0.5 mol of acetic acid and 0.5 mol of sodium acetate to one liter forms a buffer that is 0.5 molar in acetic acid and sodium acetate
    2. Partially neutralizing an acid (or base)
      Neutralizing 500 mL of 2M acetic acid with 500 mL of 1M NaOH creates a solution that is .5 molar in acetic acid and sodium acetate (see exercise 17.2.1)

    Exercise \(\PageIndex{1}\)

    What is the concentration of a solution made by neutralizing 500.0 ml of 2.0M acetic acid with 500.0 ml 1.0M sodium hydroxide, assume the volume of the solutions are additive (500ml + 500ml = 1,000ml)


    \[HC_2H_3O_2 + NaOH \rightarrow NaC_2H_3O_2 + H_2O\]

    Initial moles acetic acid = 0.5000(2.0) = 1.0 mol
    Initial moles sodium hydroxide = 0.5000(1.0) = 0.50 mol

    Final mols sodium acetate formed after complete consumption of limiting reagent (sodium hydroxide)
    \(0.50molNaOH)\left (\frac{1molNaC_2H_3O_2}{1molNaOH}\right)=0.50molNaC_2H_3O_2\)

    Final mols excess reagent (acetic acid) after complete consumption of limiting reagent
    \(1.0molHC_2H_3O_2-0.50molNaOH)\left (\frac{1molHC_2H_3O_2}{1molNaOH}\right)=0.50molHC_2H_3O_2\)

    The final concentration is the moles per liter, or \(\frac{n}{V_T}\) where the total volume is 500.0ml+500.0ml = 1.0000L.


    Note: When mixing solutions of acids and bases the concentration of the species are being changed by two different factors, chemical reaction and diluting. Even if you added water and it did not react, the concentration would go down.

    Buffer (Henderson Hasselbach) Equation

    The buffer equation is also known as the Henderson-Hasselbach equation is an expression of the equilibrium constant in terms of pValues, and making the assumption that the extent of reaction is negligible when you solve the ICE diagram for an acid or base in equilibrium with its conjugate (see section

    \[K_{a}=\frac{[H^+]_e[A^-]_e}{[HA]_e}=\frac{x([A^-]_i+x)}{[HA]_i-x}\approx\frac{x[A^-]_i}{[HA]_i} \]

    That is, we can use the initial acid and salt concentration

    \[K_{a}=\frac{[H^+]_e[A^-]_i}{[HA]_i} \label{17.2.8}\]

    One can then rearrange and solve for pH with the Henderson Hasselbach eq. Note, the analogous problem could be solved for a base and its salt, in which case you use pOH and Kb.



    There are two important characteristics of a buffer, the pH it maintains, and how much external acid or base it can handle, its capacity. The Henderson Hasselbach equation is preferred over eq 17.2.8 not only because hydronium ion concentration is typically measured in terms of pH, but also because it makes it easy to identify the pH it works at, and the buffer capacity, how much external acid or base it can neutralize,

    Buffer pH

    The ideal pH or pOH for a buffer to work at is the pK value of the acid or base that makes the buffer. Lets consider an acid and its salt, but realize we could apply this to a base and its salt. When the concentration of the acid equals the salt, the buffer can neutralize equal amounts of an external acid or base. At this point the \(log \frac{[salt]}{acid} =log1=0\) and pH=pKa .

    \[pH=pK_a+\textcolor{red}{\cancel{\log \dfrac{[salt]}{acid}}} \label{17.2.9}\]

    This can can also be seen by Equation \ref{17.2.8} where [A-] = [HA] and

    \[K_{a}=\dfrac{[H^+]_e \textcolor{red}{\cancel{[A^-]_i}}}{ \textcolor{red}{\cancel{[HA]_i}}} \label{17.2.10}\]


    Note: Equations \ref{17.2.9} and \ref{17.2.10} are for solutions where the acid and its salt are of equal concentrations, and only then is \(pH=pK_a\). To get a buffer at a different concentration the ratio of the salt to the acid (or base) needs to be adjusted.

    So the strategy for picking a buffer is choose an acid with a pKa close to the desired pH and then adjust the salt to acid ratio until they equal. Also note, if

    \[[salt]>[acid] \; then \; \frac{[salt]}{[acid]}>1\; and\; log\frac{[salt]}{[acid]} \text{is a positive number and pH is greater than pK}_a \\ [salt]<[acid] \; then \; \frac{[salt]}{[acid]}<1\; and \; log\frac{[salt]}{[acid]} \text {is a negative number and pH is less than pK}_a \]

    Note, like any logarithm, you can write this in the log form, on in the power form


    Exercise \(\PageIndex{1}\)

    Acetic acid has a KA of 1.75x10-5, which gives it a pKA of 4.76. This would make it a good acid to make a buffer with a pH of 5. What would be the acetate concentration if the acetic acid concentration is 0.10M?


    \[pH=pK_a + log \frac{[acetate]}{[acetic \: acid]} \\ \therefore \\ \frac {[acetate]}{[acetic \: acid]}=10^{pH-PK_a} \\ [acetate]=[acetic \: acid] 10^{pH-pK_a} \\ [acetate]=[0.10]10^{5-4.76}=0.174\]

    Note: Since the buffer was at a higher pH than Ka, the acetate concentration (conjugate base) is greater than the acetic acid concentration.

    Exercise \(\PageIndex{2}\)

    What would be the acetate concentration for a buffer with 0.10M acetic acid to have a pH of 10.


    For starts, pKA of acetic acid is 4.76 (exercise 17.2.2 above) and so this is a very bad choice, but lets take a look. Following the same logic as the problem above, \[[acetate]=[acetic \: acid] 10^{pH-pK_a} \\ [acetate]=[0.10]10^{10-4.76}=17,000M\]

    Note: from problem 17.2.2 the acetate concentration of a 0.10M acetic acid solution would be 17,000M, which is impossible, and so this buffer can not be made.

    Can you make a buffer with an acid that has a pH of 10, that is, with a pH of 10?

    Sure, it just means that the conjugate base is much stronger than the acid. Do exercise 17.2.3

    Exercise \(\PageIndex{3}\)

    Go to table 16.3.1 and find an appropriate acid where a 0.10 molar solution of that acid could make a buffer with a pH of 10.


    Phenol has a Ka of 1.0 x 10-10 (note the pKa value in the table does not align with that value, and 9.99 should be 10) and equal molar of phenol and its salt would work. Lets use hypoidous acid, pKa = 10.5

    \[[hypoiodite]=[hypoiodus \: acid] 10^{pH-pK_a} \\ [IO^-]=[0.10]10^{10-10.5}=0.032M\]

    Note: The table of acid ionization constants used to solve16.3.1 problem 17.2.3 had both Ka and pKa values, and if you are looking for an acid to make a buffer, looking at the pKa is much easier.

    Note: In exercise 17.2.1 the pH we wanted the buffer at was greater than Ka and the resulting salt concentration in the buffer was higher than the acid, but in problem 17.2.3 the pH we wanted was lower than the Ka and so the acid concentration of the buffer was greater than the salt.

    Note: You need to do this for bases. If asked to calculate and appropriate base and conjugate acid pair to make a buffer of a certain pH, it is probably easier to convert the question to pOH. That is, in these equations, you are dealing with pOH.

    Exercise \(\PageIndex{4}\)

    What is the base to salt ratio for a buffer of propylamine with a pH of 10?. Use table 16.3.2 to get the Kb value of propylamine


    From table 16.3.2 the pKb value for propylamine is 3.46, and a solution with a pH of 10 has a pOH of 4, so this is a reasonable choice.

    \[pOH=pK_b + log \frac{[C_3H_7NH_3^+]}{[[C_3H_7NH_2]} \\ \therefore \\ \frac{[C_3H_7NH_3^+]}{[[C_3H_7NH_2]} =10^{pOH-PK_b} \\ \text{taking the reciprocal of both sides to get the base to salt ratio} \\ \frac{[C_3H_7NH_2]}{[[C_3H_7NH_3^+]} =10^{PK_b-pOH} \\ =10^{3.46-4}=0.29\]

    Buffer Capacity

    The buffer capacity deals with how much of an external acid or base can be neutralized so effectively that the pH does not change, and this deals with the concentrations of the acid and salt in the logarithmic term of the Henderson Hasselbach equation.

    \[pH=pK_a +\log\textcolor{red}{\frac{[salt]}{[acid]}}\]

    The greater the concentration of the salt and the acid, the stronger the buffer. Consider two buffers of acetic acid

    Buffer 1: [HA] = 2.0M, [ A-]= 1.0M

    \[pH=-\log(1.75x10^{-5}) +log\textcolor{red}{\frac{[1.0]}{[2.0]}}= 4.76 -.30=4.46\]

    Buffer 2: [HA] = 0.0020M, [ A-]= 0.0010M

    \[pH=-\log(1.75x10^{-5}) +log\textcolor{red}{\frac{[0.001]}{[0.002]}}= 4.76 -.30=4.46\]

    Both buffers have exactly the same pH, but buffer one is clearly stronger and has 1 mole of an internal base and 2 moles of an internal acid to react with an external base or acid, while buffer two only has 0.001 and 0.002 moles respectively.

    Note: If \([HA]\neq[A^-]\) the strength of a buffer with respect to external acids and bases are different. For the above two buffers, there was more acid than conjugate base, and so they had a greater capacitance towards an external base than an external acid.

    Note: when [HA]=[A-],the capacity to neutralize an external acid and base are the same, and the pH=pKa. This is one of the reasons one chooses an acid whose pKa equals the pH the buffer needs to maintain.

    The concept of buffer capacity is the same for weak bases and their conjugate acids, except that you use pOH and pKb in the equations, and sometimes those get tricky as you convert them back to pH.


    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: