Calculate the pH of (i) a 0.1 M solution of NaOH, (ii) a 0.0004 M solution of methyl orange, if methyl orange has a K_a value of 3.5 × 10⁴ - Leaving Cert Chemistry - Question c - 2010

To calculate the pH of a 0.0004 M solution of methyl orange, we use the formula that relates the acid dissociation constant to the concentrations:

$K_a = 3.5 × 10^{-4} \quad (given)$

We can set up the equation:

$K_a = \frac{[H^+][A^-]}{[HA]}$

Assuming the degree of dissociation is small, we can approximate:

$[H^+] = [A^-] \approx x \quad \text{and} \quad [HA] \approx 0.0004 - x \approx 0.0004$

Thus:

$3.5 × 10^{-4} = \frac{x^2}{0.0004}$

This implies:

$x^2 = 3.5 × 10^{-4} \times 0.0004$

Calculating x:

$x = \sqrt{3.5 × 10^{-4} \times 0.0004} \approx 1.18 × 10^{-2} \approx [H^+]$

Finally, we find:

$pH = -log[H^+] = -log(1.18 × 10^{-2}) = 2.9$

The pH curve can be represented graphically with the pH on the y-axis and the volume of NaOH added on the x-axis. The axes should be labelled clearly:

• Y-Axis: pH (0 to 14)
• X-Axis: Volume of NaOH (0 to 50 cm³)

The curve typically shows a sharp rise in pH around the equivalence point (near 25 cm³ NaOH addition), transitioning from acidic to basic. It should be a sigmoidal shape with minimal changes in pH at the beginning and end, and a steep change around the middle. Each section of the graph should indicate the pH regions clearly.

The curve does not need to be drawn on graph paper, but care should be taken to maintain the shape outlined in the marking scheme.

During the titration, as NaOH is added, the pH changes significantly, particularly around the equivalence point. The characteristic steep rise in the pH curve means that the endpoint can be detected with a wide range of indicators.

Most indicators change color around a pH range of 3 to 10, which encompasses the steep slope of the pH curve. Therefore, almost any acid-base indicator would be suitable, as they will all undergo a noticeable color change within the pH range experienced during this titration.