# What is the pH of a solution with [OH^-] = 2.0 times 10^-10 M?

Apr 17, 2017

We find $p H \cong 4$

#### Explanation:

We interrogate the equilibrium:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

We know that in aqueous solution under standard conditions that:

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {K}_{w} = {10}^{-} 14$

We take ${\log}_{10}$ OF BOTH SIDES:

${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = {\log}_{10} \left({10}^{-} 14\right)$

But by definition, the logarithmic function is that power to which you raise the base to get the specific answer.........i.e. if ${\log}_{a} B = c$, then ${a}^{c} = B$.

And thus ${\log}_{10} \left({10}^{-} 14\right) = - 14$

And so ${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = - 14$

OR $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right] = 14$

But we define $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, and so...........

$14 = p H + p O H$.

And now finally, $p H = 14 - p O H$

14-(-log_10(2.0xx10^-10)=4.30

Apr 17, 2017

4.30

#### Explanation:

Right off the bat, if you don't understand the general concepts of pH and pOH, I suggest you watch this 5-minute video before proceeding to read this answer.

Recall that pH is an index for describing the concentration of hydronium ion. Mathematically, pH is described as:

$p H = - \log \left[{H}_{3} {O}^{+}\right]$

A common mistake to make in this sort of problem is to move too quickly, and take $- \log \left(2.0 \cdot {10}^{-} 10\right)$, and get your pH as 9.70. This is wrong, because you're not given a hydronium ion concentration, but a hydroxide ion concentration. Therefore, what you've calculated is really something called the pOH, not the pH.

So how do you work this out? Basically, you'll need to know that the product of the concentration of hydroxide and hydronium ions in a solution will always equal ${10}^{-} 14$ at ${24}^{o} C$ Mathematically speaking:

$\left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right] = {10}^{-} 14$

Taking the negative log of both sides of the equation yields:

$p H + p O H = 14$

Given this, there's 2 ways to do this problem. Method one is to use the 1st equation, and solve for $\left[{H}_{3} {O}^{+}\right]$. Then, you just take the negative log of that.

Method 2 (easier, in my opinion) is to solve for the pOH, and then subtract it from 14.

Either way, you should get an answer of 4.30.

Do a reality check: your concentration of hydroxide is very small. Therefore, your pOH value would be fairly large, meaning that your pH would be small. This aligns with what we got.

Hope that helps :)