Question

If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.14 m solution of this acid?

Answer 1

pH = 2.1

Let
`HA (aq)` resembles the acid; as a weak acid (a small value of
`K_(a)`)
`HA` would partially dissociate to produce protons
`H^(+)` and
`A^(-)`, its conjugate base. Let the final proton concentration (i.e.,
`[H^(+)]`) be
`x`. (Apparently
`x \ge 0`) Construct the following RICE table:

`\left\begin{array}{cccccc}\text{R}&HA(aq)&\rightleftharpoons&H^(+)(aq) &+ &A^(-)(aq)\\\text{I} & 0.14 \; \text{M} & &\\\text{C}&-x \; \text{M}& &+x \; \text{M} & & +x \; \text{M}\\E & (0.14 - x)\; \text{M} & & x \; \text{M} & &\+x \; \text{M}\end{array}\right`

By definition, (all concentrations are under equilibrium condition)

`\left\begin{array}{ccc}K_(a)&=&[H^(+)] \cdot [A^(-)] / [HA]\\&=&x^(2) /(0.14 - x)\end{array}\right`

It is given that

`K_a = 5.4 \cdot 10^(-6)`

Equating and simplifying the two expressions gives a quadratic equation; solve the equation for
`x` gives:

`x^2 = 5.4 \cdot 10^(-6) \cdot (14 - x) \\x^2 + 5.4 \cdot 10^(-6) \cdot 14 \cdot x - 5.4 \cdot 10^(-6) \cdot 14 = 0 \\x = 0.0087 \; \text{M} \; (x \ge 0)`

The pH of a solutions equals the opposite of the logarithm of its proton concentration to base 10; thus for this particular solution

`\text{pH} = -\text{ln(}[H^(+)]\text{)} / \text{ln(}10\text{)} = 2.1`