1. BUFFER

2. WHAT IS BUFFERS??? Buffer is defined as a solution that resists change in pH when a small amount of an acid or base is added or when the solution is diluted. To maintain the pH of the reaction at an optimum value.

3. Buffer solution consists mixture of weak acid and its conjugate base or weak base with it conjugate acid at predetermined concentration or ratios. That is mixture of a weak acid and its salt or a weak base and its salt.

4. Consider an acetic acid-acetate buffer. The acid equilibrium that governs the system is : Now, we have added a supply of acetate ions to the system The hydrogen ion [ ] is NO LONGER EQUAL to the acetate ion [ ] The hydrogen ion concentration is equal to :

5. Taking the (-ve) logarithm of each of this equation : Inverting the last term, it becomes (+ve)

6. This form of the ionization constant equation is called the Handerson-Hasselbalch equation Useful for calculating the pH of weak acid solution containing its salt

7. Example : Calculate the pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.10 M Sodium acetate.

8. Need to calculate the [ ] of the acid and salt in the solution. The final volume is 30 mL: So, For HOAc, 0.10 mmol/mL X 10 mL = MHOAc X 30 mL MHOAc = 0.033 mmol/mL

9. For OAcˉ, 0.10 mmol/mL X 20 mL = M OAcˉ x 30 mL MOAcˉ = 0.067 mmol/mL = 4.76 + log 2.0 = 5.06

10. BUFFERING MECHANISM For a mixture of weak acid and its salt, it can be explain as follows. The pH is governed by the logarithm of the ratio of the salt and acid pH = constant + log [A⁻] [HA] * if solution is diluted the ratio remains constant So, the pH of the solution does not change.

11. If small amount of strong acid added it will combined with an equal amount of the A⁻ to convert it to HA. HA H⁺ + A⁻ Le Chatelier’s principle dictates added H⁺ will combined with A⁻ to form HA. The change in ratio [A⁻]/[HA] is small and hence the change in pH is small. If acid added in unbuffered solution (NaCl solution) the pH will decreased markedly.

12. If small amount of strong base is added it will combined with part of HA to form an equivalent amount of A⁻. Again, change in ratio is small. Buffering capacity : amount of acid or base that can be added without causing a large change in pH. This is determine by the concentrations of HA and A⁻.

13. ↑ concentrations,↑ acid/base can tolerate Buffer capacity of a solution is defined as β = dC BOH / d pH = - dC HA / d pH dC BOH and dC HA represents the number of moles per liter of strong base or acid. For weak acid or conjugate base buffer solution of greater than 0.001 M the buffer capacity is approximate by

14. The C HA and C Aˉ represent the analytical [ ] of the acid and its salt respectively. If we have a mixture of 0.10 mol/L acetic acid and 0.10 mol/L sodium acetate, the buffer capacity is :

15. If we add solid sodium hydroxide until it becomes 0.0050 mol/L, the change in pH is : In addition to [ ], the buffering capacity is governed by the ratio of HA to Aˉ. It is maximum when the ratio is unity that is the pH = pK a

16. Example A buffer solution is 0.20 M in acetic acid and in sodium acetate. Calculate the change in pH upon adding 1.0 mL of 0.10M hydrochloric acid to 10 mL of this solution.

17. Solution : mmol HOAc = 2.0 + 0.1 = 2.1 mmol mmol OAcˉ = 2.0 – 0.1 = 1.9 mmol The change in pH is -0.05.

18. A buffer can resist a pH change even when there is added an amount of strong acid or base greater than the equilibrium amount of H⁺ or OHˉ in the buffer.

19. Same goes for the weak base and its salt. Consider the equilibrium between the base B and its conjugate (BrØnsted) acid : The logarithmic Henderson-Hasselbalch form is derived exactly as above:

21. Example : Calculate the volume of concentrated ammonia and the weight of ammonium chloride you would have to take to prepare 100 mL of a buffer at pH 10.00 if the final concentration of salt is said to be 0.200 M

22. We want 100 mL 0f 0.200 M NH 4 Cl. Therefore : mmol NH 4 Cl = 0.200 mmol/mL × 100 mL = 20.0 mmol mg NH 4 Cl = 20.0 mmol × 53.5 mg/mmol = 1.07 × 10³ mg So, 1.07 g NH 4 Cl. Calculate [ ] of NH 3 by

23. The molarity of concentrated ammonia is 14.8 M