**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).

Looking more carefully at the titration curve of 1M acetic acid with 1M NaOH, The reaction is:

HC_{2}H_{3}O_{2} + OH^{-} C_{2}H_{3}O_{2}^{-} + H_{2}O

- The curve starts at about pH 3 for the dilute acetic acid. At this point there is approximately 99% HC
_{2}H_{3}O_{2}in the solution. - The curve first rises fairly quickly, as the free protons from the dissociated acetic acid are consumed.
- The curve then rises in a nearly linear fashion (red tangent line superposed on curve), with each hydroxide added consuming an acetic acid molecule. converting it to acetate ion. This region of the curve, where it is approximately linear, is known as the
**Buffer Region**. - When the volume of sodium hydroxide added equals 1/2 the initial volume of acid (equal moles of each species is now present since 1/2 of the acid has been consumed), the amount of acid becomes equal to the amount of salt: there is 50% HC
_{2}H_{3}O_{2}and 50% C_{2}H_{3}O_{2}^{-}.- Note that by the Henderson-Hasselbalch equation we now have pH = pK
_{a}since log 50/50 = log 1 = 0. - If we now draw a line from this point on the curve parallel to the x-axis, it will intersect the y-axis where
**pH = pK**._{a}

- Note that by the Henderson-Hasselbalch equation we now have pH = pK
- As we approach the point where equal amounts of base and acid have been added we see an increasing slope, until it reverses slope (there is an inflection) becoming almost vertical. This inflection point is the
**equivalence point**, that is the point where equal quantities (moles) of acid and base have been added to the solution. - Finally, as more base is added, the curve levels off again as we approach the pH of the pure base solution.

Syllabus / Schedule |

*© R A Paselk*

*Last modified 6 May 2013*