Find the hydrogen ion, H\(^+\) concentration and hydroxide ion, OH\(^-\) concentration in 0.06 moldm\(^{-3}\) solution of H\(_2\)SO\(_4\).
Sulphuric acid (H\(_2\)SO\(_4\)) is a strong acid that fully dissociates in water, yielding two moles of hydrogen ions (H\(^+\)) for every one mole of acid, especially in dilute solutions.
[ H\(^+\) ] = 2 X [H\(_2\)SO\(_4\)]
= 2 X 0.06
[ H\(^+\) ] = 0.12 = 1.2 X 10\(^{-1}\) moldm\(^{-3}\)
To get the [OH\(^-\)], recall that the product of the hydrogen ion concentration, [H+] and the hydroxide ion concentration ([OH−]) in an aqueous solution at standard temperature is a constant, known as the ion product of water (Kw), which is approximately1.0 x10\(^{-14}\)mol\(^2\)/dm\(^6\).
K\(_w\) = [ H\(^+\) ] [OH\(^-\)]
[OH\(^-\)] = \(\frac{K_w}{[H^+]}\)
= \(\frac{1.0 x10^{-14}}{0.12}\)
[OH\(^-\)] = 8.33 x 10\(^{-14}\)
There is an explanation video available below.
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