Question

Caluclate the pH of a solution made by mixing 45.0mL of 0.25M HA (K _a= 2.5x10 ^-5) with 15.0mL of 0.75M NaOH

Answer 1

Millimol(s) HA = (45.0 mL)(0.25 M) = 11.25 mmol.

Millimol(s) NaOH = (15.0 mL)*(0.75 M) = 11.25 mmol.

HA reacts with NaOH as per the stoichiometric equation below.

HA (aq) + NaOH (aq) -------> NaA (aq) + H₂O (l)

As per the stoichiometric equation,

1 mol HA = 1 mol NaOH.

Since there are equal number of millimol(s) of HA and NaOH, hence, the solution is neutralized and contains the weak base (A^- derived from HA) and water. Water is neutral, but A^- determines the pH of the solution.

Millimol(s) A^- formed = millimol(s) HA present = 11.25 mmol.

Total volume of the solution = (45.0 + 15.0) mL = 60.0 mL.

Molarity of A^- = (11.25 mmol)/(60.0 mL) = 0.1875 M.

A^- Establishes equilibrium in water as

A^- (aq) + H₂O (l) -----------> HA (aq) + OH^- (aq)

Since OH^- is formed, we work with K_b.

K_a of HA is given. Determine K_b of A^- by employing the relation

K_a*K_b = K_w

Where K_w = 1.0*10^-14 is the ionic product of water.

Plug in values and get

K_b = K_w/K_a

= (1.0*10^-14)/(2.5*10^-5)

= 4.0*10^-10

Write down the expression for K_b as below.

K_b = [HA][OH^-]/[A^-]

=======> 4.0*10^-10 = (x)(x)/(0.1875 – x)

Since K_b is small, assume x << 0.1875 M and hence,

4.0*10^-10 = x²/(0.1875)

=======> x² = 4.0*10^-10*(0.1875) = 7.5*10^-11

=======> x = 8.66*10^-6

[OH^-] = 8.66*10^-6 M;

pOH = -log [OH^-]

= -log (8.66*10^-6 M)

= 5.062

It is known that

pH + pOH = 14

Therefore,

pH = 14 – pOH

= 14 – 5.062

= 8.938

≈ 8.94

The pH of the solution is 8.94 (ans).