Question

What is the pH of the solution if o.017 mol of HCl is added to a buffer containing 0.015 mol of RCOOH and 0.039 mol of RCOONa (source of RCoo)? Ka = 2.1E-5 Check Oxalic acid HOOCCOOH, is a diprotic acid with K_a1 = 0.056 and K_a2 = 1.5 times 10^-4. Determine the equilibrium [HOOCCOOH] in a solution with an initial concentration of 0.15 M oxalic acid. HOOCCOOH(aq) = HOOCCOO (aq) + H^+ (aq) HOOCCOO (aq) = OOCCOO (aq) + H^+ (aq)

Answer 1

(a)

Given:

K_a = 2.1*10^-5, so pK_a = -log(K_a) = 4.67

Using the following equation:

pH = pK_a + log(moles of salt/moles of acid)

After addition of HCl, net moles of acid increase and moles of salt decrease.

So, at equilibrium:

moles of acid = 0.015 + 0.017 = 0.032

moles of salt = 0.039 - 0.017 = 0.022

Putting values in the above equation, we get:

pH = 4.67 + log(0.022/0.032) = 4.5

(b)

For the diprotic acid, the second dissociation constant is almost negligible, so the [HOOCCOOH] dissociates into the solution almost completely through the first reaction.

Consider the reaction below:

H₂C₂O₄ ----> H⁺ + HC₂O₄^-

Initial 0.15 0 0

Eqb 0.15-x x x

K_a1 = ([H⁺][HC₂O₄^-]) / [H₂C₂O₄] = x²/(0.15-x) = 0.056

Solving above equation, we get:

x = 0.0678

So, at eqb.,

[HOOCCOOH] = 0.15 - 0.0678 = 0.0822 M