Question

A buffer that contains 0.18 M of an acid, HA and 0.49 M of its conjugate base A-, has a pH of 3.71. What is the pH after 0.041 mol of HCl are added to 0.79 L of the solution?

Answer 1

Sol :-

Using Henderson-Hasselbalch equation,

pH = pK_a + log(Conjugate base/weak acid)

3.71 = pK_a + log(0.49/0.18)

3.71 = pK_a + 0.4349

pK_a = 3.275

Number of moles of HA = Molarity x Volume in L = 0.18 M x 0.79 L = 0.1422 mol

Number of moles of A^- = 0.49 M x 0.79 L = 0.3871 mol

and

Given Number of moles of HCl = 0.041 mol

ICF table is :

.........................A^- .........+...........HCl ---------------> HA...............+...................Cl^-  

Initial (I)...........0.3871 mol.........0.041 mol.............0.1422 mol......................

Change (C).......-0.041 ..............-0.041.........................+0.041.........................

Final (F) ..........0.3461 mol..........0.0 mol.....................0.1832 mol .......................

Using Henderson-Hasselbalch equation,

pH = pK_a + log(Conjugate base/weak acid)

pH = 3.275 + log 0.3461 / 0.1832

pH = 3.275 + log 1.8892

pH = 3.275 + 0.276

pH = 3.55