Question

2.) Consider a 1.00L of buffer that is 0.139M in lactic acid (C₃H₅O₃H) and 0.134M in sodium lactate (C₃H₅O₃^- Na⁺). Calculate the pH of the buffer after adding 0.0500 mol of HCl into the buffer solution. (Assume no volume change) pKa of lactic acid is 3.86

Answer 1

We know, Moles = Molarity (M) x Volume (V in L) So Moles HC3H503 = 0.1390 * 1.000 mol = 0.1390 mol Molarity of HC3H503 = 0.139 M Molariy of Lacate C3H503- = 0.134 Volume of Buffer = 1.0 L M So Moles C3H503- = 0.1340 * = 1.000 mol 0.1340 mol Initial pH of solution: Henderson equation for buffer solution pH=pKa + log [Base/Acid] here the Base is C3H503- and acid is HC3H503 pH = pka + log [C3H503-/HC3H503] = 3.860 + log [ 0.1340 0.1390 ] = 3.860 + log [ 0.9640 ] = 3.860 + -0.016 = 3.84 H+ addition to buffer : When H+ is added to the buffer, it reacts with base C3H503- and produce HC3H503. So during the addition of H+ to the buffer; C3H503- Conc/moles will decrease and HC3H503 moles / conc will increase as shown in the below table. so Moles of H+ added = 0.050 mol C3H503- (aq) + H+ (aq) <-----> HC3H503 (aq) Initial 0.1340 0.0500 0.139 Change - - 0.05 -0.05 + 0.0500 Equilibrium 0.0840 0 0.1890 ] pH = pka = 3.860 = 3.860 = 3.860 3.51 + log (C3H503- / HC3H503] + log [ 0.0840 0.1890 + log [ 0.4444 ] + -0.352