Buffer Solutions

Titration of weak acid by a strong base

Skills to develop

Adding an acid or a base to a buffer solution is the same as carrying out a titration. We discuss the pH changes in a buffer solutions using titration of weak acid by a strong base.

Buffer Solutions

Buffer solutions contain a weak acid and its salt or a weak base and its salt. The pH values of these solutions do not change much when a little bit of acid or base is added.

The blood is a natural buffer, and so are other body fluids and plant fluids due to mixtures of weak acids and bases present in them.

On this page, we explore the reasons why the pH of buffer solutions resists to change.

Buffer solutions are required for many chemical experiments. They are also useful to standardize pH meters. Thus, there are many suppliers of buffer solutions.

  • Tedia Buffer Solutions.
  • HF Scientific, Inc pH Buffers.
  • Sensorex Color coded buffers.

    There are also computer programs available to help design and make buffer solutions on the internet. For example:

  • Buffer Maker using the Henderson-Hasselbalch equation.
  • SIS Scientific Software buffer maker.

    Titration of a Weak Acid by a Strong Base

    You have investigated how the pH varies in a strong-acid and strong-base Titration.

    We illustrate the titration of a weak acid by a strong base using the following examples.

    During the titration process and before the equivalent point is reached, some acid has been neutralized by the strong base, and the solution contains a weak acid and its salt. The solution acts as a buffer.

    Example 1.

    What is the pH of a Ca M acid solution whose acid dissociation constant is Ka?
    (What is the pH of a Ca M weak acid before starting the titration?)

    Solution
    Let HA represent the weak acid, and assume x M of it is ionized. Then, the ionization and equilibrium concentration is

     HA = H+ + A-
    Ca-x   x    x
    
           x2
    Ka = ------
          Ca-x
    
    x2 + Kax - CaKa = 0
    
        -Ka + (Ka2 + 4 CaKa)1/2
    x = ---------------------
                2
    
    pH = -log(x)
    

    Discussion
    The method has been fully discussed in Weak acids and bases equilibrium. Symbols are used here, but approximations may be applied to numerical problems.

    Example 2.

    Let us make a buffer solution by mixing Va mL of acid HA and Vs mL of its salt NaA. For simplicity, let us assume both the acid and the salt solutions have the same concentration C M. What is the pH of the so prepared buffer solution? The acid dissociation constant is Ka.

    Solution
    After mixing, the concentrations Ca and Cs of the acid HA and its salt NaA respectively are

    Ca = C Va / (Va+Vs)
    Cs = C Vs / (Va+Vs)
    Assume x M of the acid is ionized. Then, the ionization and equilibrium of the acid is shown below, but the salt is completely dissociated.
     HA = H+ + A-
    Ca-x  x    x
    
    NaA = Na+ + A-
          Cs    Cs
    
    Common ion [A-] = x+Cs
    
         x(x+Cs)
    Ka = --------
         Ca-x
    
    x2 + (Ka+Cs)x - Ca Ka = 0
    
        -(Ka+Cs) + ((Ka+Cs)2 + 4 Ca Ka)1/2
    x = -----------------------------
             2
    
    pH = -log(x)
    

    Discussion
    The formulas for x and the pH derived above can be used to estimate the pH of any buffer solution, regardless how little salt or acid is used compared to their counter part.

    When the ratio Ca / Cs is between 0.1 and 10, the Henderson-Hasselbalch equition is a convenient formula to use.

         [H+] [A-]
    Ka = ----------
           [HA]
    
                    [A-]
    pKa = pH - log (----)
                    [HA]
    
    The Henderson-Hasselbalch equation is
                    [A-]
    pH = pKa + log (----)
                    [HA]
    
                    [Cs] + x
       = pKa + log (----)
                    [Ca] - x
    
    Because when x is insignificant in comparision to Cs, [A-] = Cs and [HA] = Ca

    Base
    added
    [H+]pHSee
    0.0 mL0.003162.500 A.
    0.1 9.07e-43.042 B.
    1.0 9.07e-54.042 C.
    5.0 1.0e-55.000 D.
    10.010-11.34911.349E.
    Example 3

    Plot the titration curve when a 10.00 mL sample of 1.00 M weak acid HA (Ka = 1.0e-5) is titrated with 1.00 M NaOH.

    Solution

    1. Because the concentration is high, we use the approximation
      [H+] = (CaKa)1/2
        = 0.00316
      pH = 2.500

      Note the sharp increase in pH when 0.1 mL (3 drops) of basic solution is added to the solution.

    2. When 0.1 mL NaOH is added, the concentration of salt (Cs), and concentration of acid Ca are:
      Cs = 0.1*1.0 M / 10.1 = 0.0099 M
      Ca = 9.9*1.0 M / 10.1 = 0.98
        HA = H+ + A-
       Ca-x  x    x
      
       [A-] = x  + 0.0099 (= Cs)
      
           x  (x  + 0.0099)
      Ka = -------------- = 1e-5
             0.98 - x 
      
      x2 + 0.0099 x = 9.8e-6 - 1e5 x
      x2 + (0.0099 + 1e5) x - 9.8e-6 = 0
      
      x  = (-0.0099 + (0.00992 + 4*0.98*1e-5)1/2) / 2
        = 0.000907
      
      pH = 3.042
      

      Note that using the Henderson-Hasselbalch equation will not yield the correct solution. Do you know why?

    3. When 1.0 mL NaOH is added, concentration of salt,
      Cs = 1.0*1.0 M / 11.0 = 0.0901 M
      Ca = 9.0*1.0 M / 11.0 = 0.818
        HA = H+ + A-
       Ca-x   x     x+0.0901 
      
           x  (x  + 0.0901)
      Ka = -------------- = 1e-5
             0.818 - x 
      
      x  = (-0.0901 + (0.09012 + 4*0.818*1e-5)1/2) / 2
        = 0.0000907    (any approximation to be made?)
      
      pH = 4.042
      
      Using the Henderson Hasselbalch equation yield
                      0.0901(=[A-])
      pH = pKa + log (------) = 5 - 0.958 = 4.042 (same result)
                      0.818(=[HA])
      

    4. When 5.0 mL NaOH is added, concentration of salt,
      Cs = 5.0*1.0 M / 15.0 = 0.333 M
      Ca = 5.0*1.0 M / 15.0 = 0.333 M
        HA = H+ + A-
       Ca-x  x    x+0.3333
      
           x  (x  + 0.333)
      Ka = -------------- = 1.0e-5
             0.333 - x 
      
      x 2 + (0.333-1e-5)x  - 0.333*1e-5 = 0
      
      x  = (-0.333 + (0.3332 + 4*3.33e-6)1/2) / 2
        = 0.0000010
      
      pH = 5.000
      

      Note: Using the Henderson Hasselbalch equation yield the same result

                      0.333(=[A-]
      pH = pKa + log (----) = 5 + 0.000 = 5.000
                      0.333(=[HA])
      

    5. When 10.0 mL NaOH is added, concentration of salt,
      Cs = 10.0*1.0 M / 20.0 = 0.500 M
      Ca = 0.0*1.0 M / 20.0 = 0.000
      At the equivalent point, the solution contains 0.500 M of the salt NaA, and the following equilibrium must be considered:
        A- + H2O = HA + OH-
       Cs-x        x     x Equilibrium concentrations
           [HA] [OH-] [H+]
      Kb = ---------- ----
             [A-]     [H+]
      
           Kw    1e-14            x 2
         = --- = ----- = 1e-9 = ------
           Ka    1e-5            Cs-x 
      
      x  = (0.500*1.0e-9)1/2 = 2.26e-5
      
      pOH = -log x  = 2.651; pH = 14 - 2.651 = 11.349
      

      Note: The calculation here illustrate the hydration or hydrolysis of the basic salt NaA.

    Discussion..
    Sketch the titration curve based on the estimates given in this example, and notice the points made along the way.

    Confidence Building Question

    cchieh@uwaterloo.ca