Question

9. A buffer solution was prepared by mixing 12.0 mL of 0.12 mol L lactic acid (HLac) with 18.0 mL of 0.16 mol L sodium lactate (NaLac). Kallactic acid) = 1.4 x104. (10 marks) a) Calculate the pH of the solution. Write all relevant chemical equations. State and verify any assumptions you make. 1/6] b) Two 200-ml buffer solutions were prepared with the same pH using different pairs of compounds, as outlined in the table below. Which of these two solutions then has the higher/better buffer capacity? Circle your choice and briefly explain. You DO NOT need to do any calculations. 1/4 BUFFER 1 6* 10 mol HA 6x 10 mol NaA dissolved in a total volume of 200 mL BUFFER 2 5 x 10 mol HY 2 x 10* mol Nai dissolved in a total volume of 200 ml

Answer 1

b. Buffet capacity majorly depends on the ratio of buffer components concentrations. i.e

pH = pKa + log (salt) /(acid)

The highest buffer capasity (at a given total concentration of Csalt+ Cacid) occures, when (salt) /(acid) = 1 i.e pH = pka

Among two buffer solutions, in the given table, the first buffer solution is having the value of ratio is 1.

Thus among two given buffer solutions, the first buffer solution is having highest buffer capacity.

The buffer solution of Lartic aid and sodices lactate es SHS og in total solution. Now HH503 + Cats 0₃ As the solution is dilute, we have to calculate, the concentration of HGH5O3 and from given question, total volume = do ml + 18mL 300L 8 10 L = 0.03 L • putting the formala, Neva = Naya Fore HGH503 , (o.sa moil")0.04a ) = MG (0.031) Na (Lactic acid) Fore C3H503, (0.16 701 L") C0.018 L) > Na (Nalac) -0.096 MOIL! Henderson's Isalt] pH pka + log Taced] Ах |о [0.048 mol L = 0.048 mocht Na COBL) Novet platting Equation, Н. IL] Il -Iogle.4 x 104) + log [ 8.896 mol 11 A-lag 11+ loga Н P = 4.154
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