Calculate the concentration of the \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) ions in an aqueous solution of pH \(5.0 .\)

Short Answer

Expert verified
The concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is \(10^{-5.0}\) M and the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-9}\) M

Step by step solution

01

Compute Concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\)

The definition of pH is \(-\log[H_{3}O^{+}]\). So, to find the concentration of the hydrogen ion \([H_{3}O^{+}]\), we can rearrange this equation and use the given pH of 5.0 to find \(H_{3}O^{+}\) = \(10^{-pH}\) = \(10^{-5.0}\) M.
02

Calculate the Concentration of \(\mathrm{OH}^{-}\) ions

Now, to find the concentration of hydroxide ions \([OH^{-}]\), we use the ion product of water, \(Kw = [H_{3}O^{+}][OH^{-}]\). The value of \(Kw\) at 25 degrees Celsius is \(1.0 \times 10^{-14} M^2\). By substituting the value of \([H_{3}O^{+}]\) calculated in step 1 into the \(Kw\) equation and solving for \([OH^{-}]\): we get \([OH^{-}] = \frac{Kw}{[H_{3}O^{+}]}\) = \(\frac{1.0 \times 10^{-14}}{10^{-5.0}}\)= \(1.0 \times 10^{-9}\) M.

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