# Question #1f9af

Apr 27, 2016

$6.78$

#### Explanation:

pH 6 refers to a hydrogen ion concentration of ${10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}$.

If this is diluted by a factor of 100 then:

$\left[{H}_{\left(a q\right)}^{+}\right] = {10}^{- 8} \textcolor{w h i t e}{x} \text{mol/l}$

However water dissociates:

${H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

For which:

${K}_{w} = \left[{H}_{\left(a q\right)}^{+}\right] \left[O {H}_{\left(a q\right)}^{-}\right] = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^(2)."l}}^{- 2}$ at ${25}^{\circ} \text{C}$.

This means that at ${25}^{\circ} \text{C}$:

$\left[{H}_{\left(a q\right)}^{+}\right] = {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}$

So the total concentration of ${H}_{\left(a q\right)}^{+}$ is:

${10}^{- 8} + {10}^{- 7} = 1.1 \times {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}$

However this will not be the total ${H}^{+}$ ion concentration at equilibrium.

You can imagine some pure water where the ${H}_{\left(a q\right)}^{+}$ concentration is ${10}^{- 7} M$.

Now we add some acid $H X$ such that its concentration is ${10}^{- 8} M$. We have now disturbed a system which was at equilibrium.

According to Le Chatelier the system will respond by opposing this change and shift to the left.

We can show this in an ICE table based on concentrations:

$\text{ "H_2O" "rightleftharpoons" "H^+" "+" } O {H}^{-}$

$\textcolor{red}{I} \text{ " -" "1.1xx10^(-7)" } {10}^{- 7}$

$\textcolor{red}{C} \text{ "-" "-x" } - x$

$\textcolor{red}{E} \text{ "-" "(1.1xx10^(-7)-x)" } \left({10}^{- 7} - x\right)$

So we can write:

$\left(1.1 \times {10}^{- 7} - x\right) \left({10}^{- 7} - x\right) = {10}^{- 14}$

This becomes:

${x}^{2} - \left(2.1 \times {10}^{- 7}\right) x + 0.1 \times {10}^{- 14} = 0$

This is a quadratic equation which can be solved for $x$. I won't go into it here but:

$x = 5.0 \times {10}^{- 9} \textcolor{w h i t e}{x} \text{mol/l}$

(Ignoring the absurd root.)

From the ICE table we get:

$\left[{H}_{\left(a q\right)}^{+}\right] = \left(1.7 \times {10}^{- 7} - x\right)$

$\therefore \left[{H}_{\left(a q\right)}^{+}\right] = \left(1.7 - 0.05\right) \times {10}^{- 7} = 1.65 \times {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}$

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right] = - \log \left[1.65 \times {10}^{- 7}\right] = 6.78$

This is an example of "The Common Ion Effect".

We would expect this result as the pH will tend towards 7 as the solution becomes more dilute.