Question

Question 2: Calculate the pH of 100 mL of a buffer solution that is 0.063 M in phenol, C6H5OH (Ka = 1.3 x 10-10), and 0.13 M in NaC6H50 after (a) 0 mL and (b) 10 mL of 0.75 M HCl is added

Answer 1

To find the initial pH of the Buffer Concnetration of PhoH Concnetration of Pho- Volume of Buffer = 0.0630 mol L-1 0.1300 mol L-1 0.10 L (100 ml = 100 ml * 0.001 L/1.0 ml = 0.1L) Moles of acid or Base So, Moles of PhoH Moles of Pho- = = = Molarity (M) * Volume (V in L) 0.063 mol L-1* 0.100 L 0.130 mol L-1* 0.100 L = = 0.0063 0.0130 mol mol Ka of PhoH The Realation B/w Ka & Pka is; pka pka = = = = 1.30E-10 -log ka -log ( 1.300E-10 ] 9.886 PhoH is a weak acid and Pho is the conjugate base of weak acid. So both together act as Buffer solution. To find the pH of Buffer solution Henderson equation is r Con. Base pH = pka log L Acid + J pH = pka + 9.886 + Pho- log L PhoH ] r 0.013000 log L 0.00630 log [ 2.0635 ] 0.3146 + + 9.886 9.886 10.201 10.20 pH of buffer solution at zero ml Acid Addition To find the pH after addition 10 ml acid to the buffer Moles of Acid added = Molarity (M) * Volume (V in L) Molarity of Acid = 0.750 mol L-1 Volume of Acid = 10.00 ml * 0.001 L/1.0 mL 0.010 L So, Moles of [H+] added = 0.750 mol L-1 * 0.010 L 0.0075 mol When Acid is added to the buffer, it reacts with the Base component ( Pho-) and produce its conjugate Acid (PhOH). So during the addition of strong acid to the buffer, Pho- moles are decreased and PhoH moles are increased as shown in the below table. The Net ionic equation after addition of the strong base to the Buffer is Pho- (aq) + H+ (aq) {-----> Phoz (aq) Initial 0.0130 0.0075 0.0063 Change -0.0075 -0.0075 + 0.0075 Equilibrium 0.0055 0.000 0.0138 Henderson equation for buffer solution is pH = pka + log | Acid Here the Base is Pho- and Acid is PhoH 0.0055 pH = 9.886 + log 0.0138 ) 9.886 + log [ 0.3986 ] 9.886 -0.3995 = 9.487 | Conj. Base ] + + pH of buffer after addition of acid = 9.49