What is the pH of a 3.9*10^-8 M NaOH solution?

Jul 7, 2016

The pH is 7.08.

Explanation:

The base is so dilute that we must also consider the $\text{OH"^"-}$ that comes from the autoionization of water.

$\text{H"_2"O" ⇌ "H"^+ + "OH"^"-}$

K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"

We start by noting that charges must balance.

If the $\text{OH"^"-}$ comes from $\text{NaOH}$, then

$\left[{\text{OH"^"-"] = ["H"^+] + ["Na}}^{+}\right]$

The ${\text{H}}^{+}$ comes from the autoionization of water, so

$\left[\text{H"^+] = (1.00 × 10^"-14")/["OH"^"-}\right]$

Also, ["Na"^+] = 3.9 × 10^-8"

["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"

$\left[\text{OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-}\right]$

["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0

["OH"^"-"] = 1.21 × 10^"-7"

"pOH" = "-log"(1.21 × 10^"-7") = 6.92

$\text{pH" = "14.00 - pH" = "14.00 - 6.92} = 7.08$

Jul 10, 2016

$\textsf{p H = 7.08}$

Explanation:

We would predict that as this is such a low concentration the $\textsf{p H}$ will be close to 7.

This is an incredibly low concentration, so we must take into account the $\textsf{O {H}^{-}}$ ions which arise from the auto - ionisation of water:

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

For which:

sf(K_w=1xx10^(-14)" ""mol"^(2)."l"^(-2)  at sf(25^@"C".

This means that the concentration of sf(OH^- ions is $\textsf{{10}^{- 7} \text{M}}$ arising from the water.

A tiny amount of $\textsf{O {H}^{-}}$ ions is added such that the concentration is sf(3.9xx10^(-8)"M".

It might be reasoned, therefore, that the total $\textsf{O {H}^{-}}$ ion concentration will be sf((3.9xx10^(-8))+10^(-7)"M".

However this will not be the $\textsf{O {H}^{-}}$ ion concentration at equilibrium.

By adding those ions we have disturbed a system which is at equilibrium.

Le Chatelier's Principle tells us that the system will respond by opposing that change, thus restoring the equilibrium which represents the lowest energy state.

The reaction quotient $\textsf{Q}$ is given by:

$\textsf{Q = \left[{H}_{\left(a q\right)}^{+}\right] \left[O {H}_{\left(a q\right)}^{-}\right] = {10}^{- 7} \times \left[\left(3.9 \times {10}^{- 8} + {10}^{- 7}\right)\right]}$

$= \textsf{1.39 \times {10}^{- 14} {\text{ ""mol"^2."l}}^{- 2}}$

This shows that $\textsf{Q > K}$ which confirms that the equilibrium will shift to the left. More $\textsf{{H}^{+}}$ will be consumed by the extra $\textsf{O {H}^{-}}$ ions to produce more water.

To calculate the $\textsf{p H}$ we need to get the concentration of $\textsf{{H}^{+}}$ ions at equilibrium which we can do by setting up an ICE table:

" "sf(H_2O" "rightleftharpoons" "H^(+)" "+" "OH^(-))

$\textsf{\textcolor{red}{I} \text{ "--" "10^(-7)" } 1.39 \times {10}^{- 7}}$

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

$\textsf{\textcolor{red}{E} \text{ "--" "(10^(-7)-x)" } \left(1.39 \times {10}^{- 7} - x\right)}$

So:

$\textsf{{K}_{w} = \left({10}^{- 7} - x\right) \left(1.39 \times {10}^{- 7}\right) = {10}^{- 14}}$

This simplifies to:

$\textsf{{x}^{2} - \left(2.39 \times {10}^{- 7}\right) x + 0.39 \times {10}^{- 14} = 0}$

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

Discarding the absurd root we get:

$\textsf{x = 0.1765 \times {10}^{- 7} \text{mol/l}}$

So:

$\textsf{\left[{H}_{e q m}^{+}\right] = \left(1 \times {10}^{- 7} - 0.1765 \times {10}^{- 7}\right) = 0.824 \times {10}^{- 7} \text{mol/l}}$

sf(pH=-log[H_(eqm)^+]=-log[0.824xx10^(-7)]

$\textsf{\textcolor{red}{p H = 7.08}}$

You may see this more generally described as "The Common Ion Effect" . $\textsf{O {H}^{-}}$being the common ion in question.