Chemistry help??? Find the Change in Ph?

A beaker with 130 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.60 mL of a 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.



Can someone help me solve this??Chemistry help??? Find the Change in Ph?
Let's assume that the salt in the buffer is sodium acetate, NaAc. There are two processes at work



HAc 鈫?H+ + Ac- and NaAc 鈫?Na+ + Ac-



where H+ represents the hydronium ion and Ac- the conjugate base. The HAc is a weak acid, so equilibium in the first is well over to the left, that is, the degree of dissociation is small. On the other hand, it may be assumed that the salt is completely dissociated.



As there is a volume change when the hydrochloric acid is added, we shall carry out much of the calculation in terms of the amount of each component, ie the number of moles present, rather than the concentration. The amount of acetate present, either as undissociated acid or conjugate base, is constant throughout the events described, and amounts to (130/1000) x 0.100 = 0.013 mol. Since we are told that the pH of the buffer is 5.00, we can work out the ratio of the concentrations of these from the Henderson-Hasselbalch equation



pH = pKa + log[Ac-]/[HAc]



or [Ac-]/[HAc] = 10^(5.00 - 4.76) = 1.738



and since [Ac-] + [HAc] = 0.100 M, this gives



[Ac-] = 6.348 x 10^(-2) M, so number of moles of Ac- in buffer is

(130/1000) x 6.348 x 10^(-2) = 8.252 x 10^(-3) mol



[HAc] = 3.652 x 10^(-2), so number of moles of HAc in buffer is

(130/1000) x 3.652 x 10^(-2) = 4.748 x 10^(-3) mol



When the hydrochloric acid is added to the buffer, the H+ ions from it combine with Ac- ions to form additional HAc, whose concentration increases while that of the conjugate base decreases by the same amount, since the sum total is constant. As [H+] is small in comparison to both [HAc] and [Ac-], and does not change greatly (since the solution is a buffer), it is a reasonable approximation to assume that this process goes to completion. Since the number of moles of H+ in the HCl added is (6.60/1000) x 0.330 = 2.178 x 10^(-3) mol



final amount of HAc = 4.748 x 10^(-3) + 2.178 x 10^(-3) = 6.926 x 10^(-3) mol



final amount of Ac- = 8.252 x 10^(-3) - 2.178 x 10^(-3) = 6.074 x 10^(-3) mol



Hence the final pH, from the Henderson-Hasselbalch equation, is



pH = pKa + log[6.074 x 10^(-3)/6.926 x 10^(-3)] = 4.760 - 0.057 = 4.703



Note that, since these amounts are both dissolved in the same volume of solution, we do not need to convert them from moles to molarities, the ratio being the same.