8. (a) (i) Write an expression for the self-ionisation of water - Leaving Cert Chemistry - Question 8 - 2008

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

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

At 25 °C, the value of Kw is given as 1.0 × 10⁻¹⁴. To show that the pH of pure water is 7.0, we use the fact that in pure water, the concentrations of hydrogen ions and hydroxide ions are equal:

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

Thus, we have:

$K_w = x^2 = 1.0 imes 10^{-14}$

Solving for x gives:

x = rac{1.0 imes 10^{-14}}{1.0} = 1.0 imes 10^{-7}

The pH is calculated as follows:

$pH = - ext{log}[H^+] = - ext{log}(1.0 imes 10^{-7}) = 7.0$

For a strong monobasic acid, the concentration of hydrogen ions directly equals the concentration of the acid. Thus, for a 0.5 M solution:

$[H^+] = 0.5$

The pH can be calculated using:

\approx 0.3 $$

For a weak acid, we use the formula:

$pH = - ext{log}(K_a imes M)$

Substituting the given values:

= - ext{log}(9.0 imes 10^{-6}) \ \approx 5.0 $$ Thus, the pH is approximately 5.0.