Question

As the weak acid (HA) is titrated with a strong base (OH−), the neutralization reaction is:

HA + OH− → A− + H2O.
As the neutralization occurs, [HA] decreases and [A−] increases, allowing the solution to act as a buffer. Suppose 0.0015 mol HA and 0.0005 mol A− are present in 125 mL solution.
(a) What are the approximate concentrations of HA and A−?
(b) If the pKa of HA is 3.90, what is the pH of this solution?

Answer 1

a) [HA] = 0.012 M; [A-] = 4 E-3 M

b) pH = 3.47

Step-by-step explanation:

neutralization reaction:

• HA + OH- → A- + H2O

∴ mol HA = 0.0015 mol

∴ mol A- = 0.0005 mol

∴ Vsln = 125 mL = 0.125 L

a) approx. concentration:

⇒ [HA] = 0.0015 mol / 0.125 L = 0.012 mol/L

⇒ [A-] = 0.0005 mol / 0.125 L = 4 E-3 mol/L

b) pH = ?

∴ pKa = 3.90 = - Log Ka

⇒ Ka = 1.26 E-4 = ([H3O+]*( 4 E-3 + [H3O+])) / (0.012 - [H3O+])

⇒ 1.512 E-6 - 1.26 E-4[H3O+] = 4 E-3[H3O+] + [H3O+]²

⇒ [H3O+]² + 4.126 E-3[H3O+] - 1.512 E-6 = 0

⇒ [H3O+] = 3.3866 E-4 M

⇒ pH = 3.4702