Question

Calculate the percentage error in hydronium ion concentration made by neglecting ionisation of water in 10$^{-6}$ M NaOH.

Answer 1

Hint: When there is dissociation of hydrogen ions, it leads to the formation of hydronium ion., and hydroxide ion. Calculate the concentration of hydronium ion, considering the ionisation of water, and neglecting the ionisation of water. The % error would be calculated.


Complete step by step solution: Firstly, it is given that we have a solution 10$^{-6}$ M NaOH. So, we will calculate the pH of it, as we know this solution is formed using the water.
Thus, let us consider two cases; in the first one will have the ionisation, but in the second without ionisation of water.
Now, if we neglect the ionisation of water, then concentration of [OH$^{-}$] ion will be equal to the concentration of NaOH i.e. [OH$^{-}$] = 10$^{-6}$ M
As we know, the concentration of hydronium ion, and the concentration of hydroxide ion is equal to the equilibrium constant of water i.e. 10$^{-14}$, or
[H$_3$O$^{+}$][ OH$^{-}$] = 10$^{-14}$,
Then, [H$_3$O$^{+}$] = 10$^{-14}$/ 10$^{-6}$ = 10$^{-8}$ M = 10 $\times$ 10$^{-9}$ M
Now, if we consider the ionisation of water, then water will be ionised i.e.
2H$_2$O $\rightleftharpoons$ H$_3$O$^{+}$ + OH$^{-}$
The concentration of hydronium ion is equal to the concentration of hydroxide ion, but in this case we have a base NaOH, it will also ionise i.e.
NaOH $\rightarrow$ Na$^{+}$ + OH$^{-}$
Now, proceeding with the calculations; x is the concentration of hydronium ion including the ionisation of water
2H$_2$O $\rightleftharpoons$ H$_3$O$^{+}$ + OH$^{-}$
                                                      x 10$^{-6}$+x
(10$^{-6}$+x represents the concentration of hydroxide ion depicting the common ion effect)
K$_w$ = 10$^{-14}$ = x $\times$ (10$^{-6}$+x), K$_w$ represents the equilibrium constant of water.
Thus, x = 9.9$\times$ 10$^{-9}$ M
Let us now calculate the percentage error, i.e. % error = $\dfrac{10 \times 10^{-9} - 9 \times 10^{-9} }{9.9 \times 10^{-9}}$$\times$ 100
Therefore, % error = 1%
In the last we can conclude the percentage error in the hydronium ion by neglecting the ionisation of water is 1%.


Note: Don’t get confused while considering the concentration of hydroxide ion. In this reaction both the concentrations will be considered due to the common ion effect, and the equilibrium of ionisation of water will be shifted backward, because the concentration of hydroxide ion is increasing in case of NaOH.