# Which Solution Below Has The Highest Concentration Of Hydroxide Ions

For example, 1) pH of 12, and a pOH of 2 2) pH of 3, and a pOH of 11 Wouldn't the solution with the highest concentration of hydrogen ions be #1 and the highest concentration of hydroxide ions be #2. If not, please explain to me why. The higher the pH, the more hydroxide ions. PH measures the presence of H+ ions in solution. Since pH + POH = 14, the higher the pH, the lower the pOH and thus the higher concentration of OH in solution. This is because p means -log of. The answer is E.

## 2 Answers

#### Explanation:

Now in aqueous solution, under standard conditions, the following equilibrium operates...

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

And ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$...and if we take ${\log}_{10}$ of both sides..we get...

$14 = p H + p O H$ And so we gots... $A . \left[{H}_{3} {O}^{+}\right] = 0.001 \cdot m o l \cdot {L}^{-} 1$

$B . \left[{H}_{3} {O}^{+}\right] = 0.0001 \cdot m o l \cdot {L}^{-} 1$

$C . p O H = 12 ,$...so $p H = 2$, and thus $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1 \equiv 0.01 \cdot m o l \cdot {L}^{-} 1$

$D .$$p H = 7$ and thus $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$

$E .$$\left[H {O}^{-}\right] = {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$, $p O H = + 6$, $p H = 8$, and so $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$

And so the HIGHEST concentration with respect to ${H}_{3} {O}^{+}$ NECESSARILY demands the LOWEST CONCENTRATION with respect to $H {O}^{-}$. Note that we use ${H}_{3} {O}^{+}$ and ${H}^{+}$ interchangeably...

Capisce?

#### Explanation:

Let's examine the options by looking at their $' p H '$.

$A .$ has a $' p H ' = 3$.

For $B .$, the $' p H '$ of the nitric acid would be $- \log \left[{10}^{-} 4\right] = 4$, as it is an acid, and so the equation corresponds to the $' p H '$.

For $C .$, the $' p H '$ of the solution would be $14 - ' p O H ' = 14 - 12 = 2$.

$D .$ is pure water, and therefore has a $' p H ' = 7$.

For $E .$ the $' p O H '$ of the sodium hydroxide would be $- \log \left[{10}^{-} 6\right] = 6$, as it is a base, and so the logarithm equation responds to the $' p O H '$.

Since the $' p H '$ is related to $' p O H '$ by the equation,

$' p H + p O H ' = 14$ at ${25}^{\circ} ' C '$, we can find the $' p O H '$ of the following choices.

$A .$ has a $' p O H ' = 14 - 3 = 11$.

$B .$ has a $' p O H ' = 14 - 4 = 10$.

$C .$ has a $' p O H ' = 12$, as stated.

$D .$ has a $' p O H ' = 14 - 7 = 7$.

$E .$ has a $' p O H ' = 6$, as calculated.

Notice how,

### Which Solution Below Has The Highest Concentration Of Hydroxide Ions Formed

$6 < 7 < 10 < 11 < 12$

or

$E . < D . < B . < A . < C .$

Since $' p O H ' = - \log \left[O {H}^{-}\right]$, or $\left[O {H}^{-}\right] = {10}^{- ' p O H '}$, and so the hydroxide ion concentration is inversely proportional to the $' p O H '$, i.e. more $\left[O {H}^{-}\right]$ means less $' p O H '$, and vice-versa.

### Which Solution Below Has The Highest Concentration Of Hydroxide Ions Chegg

Since $C .$ has the highest $' p O H '$, then that means that it has the least $\left[O {H}^{-}\right]$ concentration, and so it is the answer.