# The pH of a 0.000001M solution of NaOH?

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#### Explanation

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#### Explanation:

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Oct 21, 2017

$\textsf{p H = 7.04}$

#### Explanation:

This is a very dilute solution of NaOH so I would expect the pH to be 7 or very slightly above.

Because the concentration is so low we must also take into account the ions that are formed from the dissociation of water:

$\textsf{{H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + O {H}^{-}}$

sf(K_w=[H^+][OH^-]=10^(-14) at $\textsf{{25}^{\circ} C}$

If you do a thought experiment you can imagine adding a small amount of NaOH to water such that its concentration = $\textsf{{10}^{-} 6 \textcolor{w h i t e}{x} \text{mol/l}}$.

The total concentration of $\textsf{O {H}^{-}}$ ions is therefore $\textsf{{10}^{- 6} + {10}^{- 7} = 1.1 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$.

However this is not the concentration of $\textsf{O {H}^{-}}$ ions at equilibrium so we cannot use this value to get the pH.

By adding those extra $\textsf{O {H}^{-}}$ ions we have disturbed a system which is already at equilibrium. If you are familiar with Le Chatelier's Principle you will understand that such a system will react by opposing the change applied to it.

It will do this by shifting to the left thus reducing the concentration of $\textsf{O {H}^{-}}$ ions to restore the value of $\textsf{{K}_{w}}$.

We need to set up an ICE table based on $\textsf{\text{mol/l}}$:

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

$\textsf{I \text{ "" "" "" "" "" "" "" "" "10^(-7)" "" "" } 1.1 \times {10}^{- 6}}$

$\textsf{C \text{ "" "" "" "" "" "" "" "" "-x" "" "" "" } - x}$

$\textsf{E \text{ "" "" "" "" "" "" "" "10^-7-x" "" } 1.1 \times {10}^{- 6} - x}$

$\therefore$$\textsf{\left(1.1 \times {10}^{- 6} - x\right) \left({10}^{- 7} - x\right) = {10}^{- 14}}$

Multiplying out this gives:

$\textsf{{x}^{2} - \left(12 \times {10}^{- 7}\right) x + {10}^{- 14} = 0}$

This is a quadratic equation which can be solved using the quadratic formula:

This gives:

$\frac{\textsf{x = \left(12 \times {10}^{- 7} \pm \sqrt{140 \times {10}^{- 11}}\right)}}{2}$

Ignoring the absurd root gives:

$\textsf{x = 0.0839 \times {10}^{- 7}}$

From the ICE table we get:

$\textsf{\left[{H}^{+}\right] = 1 \times {10}^{- 7} - 0.0839 \times {10}^{- 7} = 0.91607 \times {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(0.91607 \times {10}^{- 7}\right) = 7.04}$

As expected, very close to neutral.

This is why spillages of acid or alkali are irrigated with copious amounts of water to bring the pH towards neutrality.

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