Question

calculate the ph of a solution containing 0.15m acetic acid and .25M sodium acetate for acetic acid ka=1.8x10^-5 if .0010 mol HCL is added to 1.0 L buffer as in part a calculate the final ph of the buffer. show work with ICE

Answer 1

P^H of buffer is given by P^H = P^K_a + log ([salt]/ [acid])

P^K_a = -log K_a = -log (1.8 * 10^-5) = 4.74

Salt (sodium acetate) = 0.25

Acid (acetic acid ) = 0.15

Before addition of HCl, P^H = 4.74 + log (0.25/0.15)

P^H = 4.74 + (0.2218)

P^H = 4.9618

after addition of HCl, reaction is

CH₃COONa + HCl --------> CH₃COOH + NaCl

Intial moles 0.25*1 0.001 0.15 *1 (Moles = concen. * vol. (in litre))

At equilibrium 0.25-0.001 - 0.15 + 0.001

= 0.249 = 0.151

P^H for buffer again = P^K_a + log ([acetate salt] / [acetic acid])

P^H = 4.74 + log (0.249/0.151)

P^H = 4.74 + (0.217)

So after addition of HCl

P^H = 4.957