Question

Suppose that you have 155 mL of a buffer that is 0.4800 M in both propanoic acid (C2H5COOH) and its conjugate base (C2H5COO−). Calculate the maximum volume of 0.4500.450 M HCl that can be added to the buffer before its buffering capacity is lost.

Answer 1

Buffer capacity is said to be lost if the pH of the buffer changes by 1 unit. Given that conc. of buffer = 0.480 M volume of buffer = 155 mL Ka for CH3COOH = 1.3x10-5 pka = -log(1.3x10-5) = 4.89 Let's find the initial pH of the buffer solution before addition of HCI By Henderson equation, pH = pKa+log [base] [acid] pH = pka +log LCH,C00"] ° [C, H,COOH] 0.480 – 4.89 =pH = 4.89+ log1 0.480 Buffer capacity is said to lost if the pH of the solution is changed by 1 pH unit. Since we are going to add HCI, the pH of the solution will be decreased. So the pH of the solution = 4.89- 1 = 3.89 Let x be the number moles of HCl to be added to the buffer to change the pH.

Number of moles of C2H3COOH = M*V = 0.480 M * 155 mL = 74.4 mmol Number of moles of C,H,COO-= M*V = 0.480 M * 155 mL = 74.4 mmol C2H5COO + H+ C2H5COOH Initial 74.4 74.4 Add Change -X Equilibrium 74.4 -x 74.4 + x pH = pKa+log [C,H,COOH] 74.4-3 +log => 3.89 = 4.89 74.4 + x log2 74.4 - x -=-1 • 74.4+x 74.4 - x = 10-1=0.1 74.4 + x 74.4 - x=0.1+(74.4+x) 74.4 - x= 7.44 +0.1x 74.4 -7.44= 0.1x+x 1.1x=66.96 x=60.87 mmol number of moles of HCl = 60.87 mmol Molarity of HCl = 0.450 M Volume of HCI- moles 60.87 mmol = 135.27 mL =135 mL molarity 0.450M Volume of HCl to be added = 135 mL