*Acids and Bases, cont.*

Let's look at a second way to solve buffer problems, using something called the *Henderson-Hasselbalch equation*. The Henderson-Hasselbalch equation is simply a log version of the equilibrium expression for acid dissociation.

K_{a} = [H^{+}][A^{-}] / [HA], where HA is the acid and A^{-} is its salt (conjugate base)

If we take logs of both sides of the equation we get:

so

Henderson-Hasselbalch equation:pH = pK_{a}+ log([A^{-}] / [HA])(Note that an analogous equation may be written for bases: pOH = pK

_{b}+ log([B^{+}] / [BH]))

So now let's do the example from last time using the Henderson-Hasselbalch equation.

Calculate the pH of a buffer made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. K

_{a}= 1.8 x 10^{-5}(pK_{a}= 4.74).

pH = pK _{a}+ log[A^{-}] / [HA]

pH = 4.74 + log(0.0250) / (0.0125)

= 4.74 + 0.301

pH = 5.04Notice that only the ratio matters, the actual concentrations of the acid and salt are not that important for determining pH!

What is the pOH of this solution? Use relationship pH + pOH = 14.00 {from [H

^{+}][OH^{-}] = 1.0 x 10^{-14}, log[H^{+}][OH^{-}] = log(1.0 x 10^{-14}), etc.}then pOH = 14.00 - 5.04 =

8.96

**Example:** What ratio of acetate ion to acetic acid would you need to make up an acetate buffer with a pH of 5.25? pK_{a} = 4.74.

Easiest way to approach this is to use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc

^{-}

Rearranging,

log[A^{-}] / [HA] = pH - pK_{a}

Substituting,

log[A^{-}] / [HA] = 5.25 - 4.74 = 0.51

Now get rid of logs by raising both sides to power of ten:

10^{log[A-] / [HA]} = 10^{0.51}

[A ^{-}] = 3.236 [HA]

**Extra example:** How many moles of sodium acetate must be added to 0.120 moles of acetic acid and sufficient water to make a liter if we want a buffer with a pH of 4.50? pK_{a} = 4.74.

Again use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc

^{-}

pH = pK _{a}+ log[A^{-}] / [HA]Rearranging and substituting,

log[OAc

^{-}] / [HOAc] = pH - pK_{a}Substituting,

log[OAc

^{-}] / [HOAc] = 4.50 - 4.74 = - 0.24Raise both sides to power of 10:

[OAc

^{-}] / [HOAc] = 0.575[OAc

^{-}] =0.575 [HOAc][OAc

^{-}] =0.069 mol

Titration curves examine the reaction of an acid with a strong base (most common and the examples we will look at) or a base with a strong acid (these titrations will look like mirror images of the titration with base reflected around the y {pH}-axis).

**Strong acid/strong base titration:** (plot drawn on board, Zumdahl Fig 15.1, p 717) In this case we can observe the titration by simply calculating the pH at each point by assuming 100% reaction. For example, assume an initial volume of 100 mL of 0.10 M HCl titrated with 0.10 M NaOH, and calculate the pH with 20 mL additions (drawn on board from data below):

Volume NaOH added |
Moles H^{+} remaining |
[H^{+}] |
pH |

0 | 0.010 | 0.10 | 1 |

20 | 0.0080 | 0.0080/0.12 = 0.0667 | 1.18 |

40 | 0.0060 | 0.0060/0.14 = 0.043 | 1.37 |

60 | 0.0040 | 0.0040/0.16 = 0.025 | 1.6 |

80 | 0.0020 | 0.0020/0.18 = 0.011 | 1.95 |

100 | ª0 | 10^{-7} |
7 |

120 | ª0 | 10^{-11.96} |
11.96 |

140 | ª0 | 10^{-12.22} |
12.22 |

200 | ª0 | 10^{-12.70} |
12.70 |

Notice that once the acid has reacted the pH changes very rapidly. Also notice that the pH levels off at high pH. This is simply due to the fact that we are approaching pure 0.10 M NaOH, which would have a pH of 13.00.

Note that the titrtion of a strong base with a strong acid is just the reflection of the titration above, as seen in text Figure 15.2, p 717 (drawn on board).

The curve below shows the titration of 1M acetic acid with 1M NaOH. In this case the pH is no longer a simple function of the concentration of H^{+}, since the acid only partially dissociates in water. It does however react completely with any added strong base until the acid has been consumed, so adding base has little effect on pH initially.

Note the following:

- For a weak acid the curve starts at a higher pH, depending on how weak the acid is.
- The equivalence point occurs at the inflection in the curve where the moles of base added equals the initial moles of acid.
- When half of the equivalence point is reached (half of the moles acid added as base), we have acid = conjugate base ([HA] = [A
^{-}]), so the pH = pK_{a} - The near-linear slope (red tangent on the plot) is the buffer region (approximately the pK
_{a}± 1 pH unit).

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*© R A Paselk*

*Last modified 2 May 2011*