10. Answer any two of the parts (a), (b) and (c) - Leaving Cert Chemistry - Question 10 - 2016

To find the molarity, we use the formula for pH:

$\text{pH} = -\log[H^+]$

Given that the pH is 2.0, we can rearrange this to:

$[H^+] = 10^{-\text{pH}} = 10^{-2.0} = 0.01 \, M$

So, the molarity of the sulfuric acid solution is 0.01 M.

Similar to the above step, we again calculate the hydrogen ion concentration:

$\text{pH} = -\log[H^+]$

Solving for molarity:

$[H^+] = 10^{-2.0} = 0.01 \, M$

Assuming the acid is weak, we can use the dissociation constant formula to calculate its molarity. Since:

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

For a weak monobasic acid with Ka = 1.8 × 10^{-5} and [H^+] = 0.01 M, we have:

K_a = 5/9(0.5555) = 0.556.\

This leads to:

$[HA] = \frac{[H^+]^2}{K_a} = \frac{(0.01)^2}{1.8 \times 10^{-5}}\approx 0.5556 \, M$

The ionic product of water, K_w, is defined as:

$K_w = [H^+][OH^-]$

At 59 °C, K_w is given as 9.0 × 10^{-14}. This means that the concentration of hydrogen ions in pure water at this temperature can be determined using the formula for K_w.

To find the hydrogen ion concentration

$[H^+] = \sqrt{K_w} = \sqrt{9.0 \times 10^{-14}} = 3.0 \times 10^{-7} \, M$

This indicates that the hydrogen ion concentration in pure water at 59 °C is approximately 3.0 × 10^{-7} M.