1. Analytical Chemistry Neutral Titration

2. Introduction Neutral titrations are considered the most volumetric analysis titrations practiced since they feature high accuracy and precision along side being quick and easy to practice. Hence, it has been used to estimate a number of organic and inorganic substances that feature acidic or basic qualities. These titrations produce water and salt. The salt might be of a neutral effect. Therefore, the eq. point is situated in a neutral medium. It also might be of an acidic effect as a result of a weak base titration such as titrating ammonia with NH 4 Cl which forms ammonium, that contains an acid which leaves the eq. point in the acidic portion. The eq. point can also be situated in an alkaline medium like, the titration of a weak acid such as acetic acid with sodium hydroxide, then sodium acetates are formed which provide acetates of a base effect.

3. 1. Titrating a strong acid with a strong base and vice versa A. Titrating a strong acid with a strong base: For e.g. titrating HCl with sodium hydroxide. Such titration characterizes by the enormous changes in the pH units around the eq. point. HCl + NaOHNaCl + H 2 O H + + OH - H 2 O this reaction is almost complete: K eq = [H 2 O] = 1 = 1 x 10 -14 K w 1 x 10 -14

4. The Zone Principle Elements in the Titration Funnel pH Calculation Before titration HClpH = - log [HCl] Before eq. point HCl + NaClpH = - log ml HCl  M HCl – ml NaOH  M NaOH ml HCl + ml NaOH (strong acid solution R=1) At eq. point NaClpH = 7 After eq. point NaCl + NaOHpOH = - log ml NaOH  M NaOH – ml HCl  M HCl ml HCl + ml NaOH pH = 14 – pOH (strong base solution) A. Titrating a strong acid with a strong base:

6. 12 10 8 6 4 2 0 0.1 M 0.01 M 0.001 M 0.1 M 0.01 M 0.001 M Ph.Ph BcP MO NaOH (ml) pH A. Titrating a strong acid with a strong base:

7. B. Titrating a strong base with a strong acid: The Zone Principle Elements in the Titration Funnel pH Calculation Before titration NaOHpOH = - log [NaOH] Before eq. point NaOH + NaClpOH = - log ml NaOH  M NaOH – ml HCl  M HCl ml HCl + ml NaOH At eq. point NaClpH = 7 After eq. point NaCl + HClpOH = - log ml HCl  M NaOH – ml NaOH  M NaOH ml HCl + ml NaOH

8. 12 10 8 6 4 2 0 0.1 M 0.001 M 0.1 M 0.001 M HCl (ml) pH B. Titrating a strong base with a strong acid:

9. 2. Titrating a weak acid with a strong base and vice versa: A. Titrating a weak acid with a strong base: Here, we obtain eq. point when the weak acid HA converts to NaA where the anion A - works as a base: A - + H 2 O OH - + HA Therefore, the eq. point will take place in the basic portion. For e.g. titrating 100 ml of the acid HA with the concentration of 0.1 molar and the contuent dissociation K a = 10 -5, with sodium hydroxide concentration 0.1 molar. 1. pH before addition 2. Before eq. point 3. At eq. point 4. After eq. point

10. 1. pH before addition: We can determine the hydrogen ion concentration of the weak acid solution as following: [H + ] = K a C a = 10 -5 x 0.1 = 10 -3 M pH = 3.00 2. Before eq. point: This means that we are coming across a solution that contains a weak acid and its conjugate, so this solution is a buffer solution. If the amount of the added base was 10 ml that means that the produced salt: [NaA] = 10 x 0.1 = 9 x 10 -3 M 100 + 10 and the remaining acid concentration: [HA] = 100 x 0.1 – 10 x 0.1 = 0.08M 100 + 10 [H + ] = K a x C a C s = 10 -5 x 0.8 = 8.9 x 10 -5 M 9 x 10 -3 pH = 4.05 A. Titrating a weak acid with a strong base

11. After adding 50 ml of the base, half the acid would be equivalent, and therefore the acid concentration would be equal to the produced salt concentration: [NaA] = 50 x 0.1 = 5 = 0.03 M 100 +50 150 [HA] = 100 x 0.1 – 50 x 0.1 = 5 = 0.03 M 100 + 50 150 [H + ] = K a = 10 -5 pH = 5.00 which means: pH = pK a This means that the hydrogen ion concentration equals the acid contuent dissociation and that is when it is equal to the acid by 50% or what is called midpoint. A. Titrating a weak acid with a strong base

12. 3. At eq. point: The acid has transformed 100% to it’s salt NaA. Hence, the medium will be basic (because the salt is produced from the reaction between a weak acid and a strong base). [NaA] = 100 x 0.1 = 0.05 M 100 + 100 [OH - ] = k w C s = 10 -14 x 0.05 = 7.1 x 10 -6 M k a 10 -5 pOH = 5.2 pH = 14 – 5.2 = 8.8 A. Titrating a weak acid with a strong base

13. 4. After eq. point: The medium will be a strong base and will calculate the pH within it by the knowledge of the added sodium hydroxide concentration, and will neglect the effect of the basic salt because of its weak contribution ( its constant dissociation to OH - equals 10 -9 ) The following picture shows the change in pH with the titration progress. Observe how much pH changes at eq. point and its relation to the strength of the titrated acid. 12 10 8 6 4 2 0 K a = 10 -3 K a = 10 -5 K a = 10 -7 K a = 10 -9 20 40 60 80 100 120 140 160 180 200 pH The titration percentage (%) The curves of titrating weak acids with sodium hydroxide solution

14. Example 1 Titrating NaOH with CH 3 COOH NaOH + CH 3 COOH CH 3 COONa + H 2 O Before titration: weak acid solution pH = -log K a C a Before eq. point: A solution containing (CH 3 COONa + CH 3 COOH) and it is a buffer solution pH = pK a + log C s C a C s = ml NaOH  M NaOH (R= 1) ml NaOH  ml CH 3 COOH C a = ml CH 3 COOH  M CH 3 COOH - ml NaOH  M NaOH ml NaOH + ml CH 3 COOH At eq. point: A solution that contains (NaOH + CH 3 COONa) neglects the effect of the salt which is a weak base and is calculated as a NaOH solution: pOH = -log ml NaOH  M NaOH - ml CH 3 COOH  M CH 3 COOH ml NaOH  ml CH 3 COOH

15. B. Titrating a weak base with a strong acid This is the inverse curve of titrating a weak acid with a strong base. In this curve, the eq. point is in the acidic portion because the salt BHX is formed. It gives the kation BH + which is of an acidic quality. BH + + H 2 O B + H 3 O + hence, the solution will obtain an acidic feature at eq. point. The following picture shows such curves. It also shows the effect of the weak base constant dissociation value at the clarity of the end point (considering the number of pH units that change meanwhile) where the equilibrium constant of the reaction equals: k eq = k b k w To achieve the value 10 -6 required in the equilibrium constant, k b must not be less than 10 -8. Accordingly, weak bases with dissociated constants less than 10 -8 can not be titrated.

16. B. Titrating a weak base with a strong acid 12 10 8 6 4 2 0 K b = 10 -3 K b = 10 -5 K b = 10 -7 K b = 10 -9 20 40 60 80 100 120 140 160 180 200 pH The titration percentage (%) The curves of titrating weak bases with hydrochloric acid solutions MO BcP Ph.Ph.

17. Example 2 Titrating NH 3 with HCl: HCl + NH 3 NH 4 Cl(R = 1) the titration curve derivation is similar to the previous titration curve. Before titration: weak base solution: pOH = -log k b C b Before eq. point: The solution (NH 3 + NH 4 Cl) which is a buffer solution: pOH = pk b + log C s C b C s = ml HCl  M HCl ml NH 3 + ml HCl C b = ml NH3  M NH3 - ml HCl  M HCl ml NH3 + ml HCl

18. At eq. point: Salt solution NH 4 Cl (weak acid) pH = - log k w  C s k b C s = ml HCl  M HCl = ml NH3  M NH3 ml HCl + ml NH3 ml NH3 + ml HCl After eq. point: Solution (HCl + NH 4 Cl) is calculated based on the fact that it is a strong acid solution and neglects the effect of the salt NH 4 Cl. pH = -log ml HCl  M HCl - ml NH3  M NH3 ml HCl + ml NH3 Example 2

19. As we referred to before, when titrating a weak base with a strong acid, the pOH at mid- titration (at the titration of 50% of the base) is equal to pk b of the titrated base: k b = [OH - ] [NH + 4 ] [NH 3 ] pk b = pOH In conclusion, we realize the great importance of the titration curves which give us an idea about the reaction. It also helps in choosing the best evidence to recognize the end point. Example 2

20. The importance of evidence used in equilibrium titrations: The evidence Evidence range Acidic colorBasic color Methyl orange3.1 – 4.4redyellow Bromocresol purple 5.2 – 6.8yellowpurple Phenol red6.8 – 8.4yellowred phenolphthalein8.3 – 10.0colorlessred pink

21. Methyl orange evidence Color: basic yellow and acidic red Phenolphthalein Color: basic pink and the acidic medium is colorless The importance of evidence used in equilibrium titrations: