# Order these substances from highest to lowest "pH"?

## $\text{NaOH}$, $\text{HBrO}$, ${\text{NH}}_{3}$, "Sr"("OH")_2, $\text{HBr}$

Nov 13, 2016

$\text{Sr"("OH")_2(aq) > "NaOH"(aq) > "NH"_3(aq) > "HBrO"(aq) > "HBr} \left(a q\right)$

Here's what we should know:

• The stronger the base, the higher the $\text{pH}$, since it dissociates more in solution than weaker bases, thus making it more basic and raising the $\text{pH}$.

• The stronger the acid, the lower the $\text{pH}$, since it dissociates more in solution than weaker acids, thus making it more acidic and lowering the $\text{pH}$.

Now let's figure out where the acids and bases fall on the pH scale. The strong bases within this list are:

• "Sr"("OH")_2
• $\text{NaOH}$

The weak base within this list is ${\text{NH}}_{3}$.
The strong acid within this list is $\text{HBr}$.
The weak acid within this list is $\text{HBrO}$.

(You have to memorize the common strong acids and bases.)

Furthermore, the strong bases with more hydroxides per formula unit dissociate more hydroxides into solution. That means "Sr"("OH")_2 is twice as concentrated with ${\text{OH}}^{-}$ as $\text{NaOH}$ is.

So far, we thus have that pH varies as follows when these are placed into solution:

$\text{Sr"("OH")_2(aq) > "NaOH"(aq) > ? > ? > "HBr} \left(a q\right)$

We now know that $\text{HBrO}$ is a weak acid, so it must dissociate less than $\text{HBr}$, meaning that it decreases the $\text{pH}$ by less from about $7$. That means the $\text{pH}$ of $\text{HBrO"(aq) > "pH}$ of $\text{HBr} \left(a q\right)$.

Now we have:

$\text{Sr"("OH")_2(aq) > "NaOH"(aq) > ? > "HBrO"(aq) > "HBr} \left(a q\right)$

Since ${\text{NH}}_{3}$ is a weak base, it increases the $\text{pH}$ by less than $\text{NaOH}$, but since it increases the $\text{pH}$ and $\text{HBrO}$ decreases the $\text{pH}$, it means that in general the same concentration of both implies that the $\text{pH}$ is higher for ${\text{NH}}_{3} \left(a q\right)$ than $\text{HBrO} \left(a q\right)$.

Therefore, our result is:

$\boldsymbol{\text{Sr"("OH")_2(aq) > "NaOH"(aq) > "NH"_3(aq) > "HBrO"(aq) > "HBr} \left(a q\right)}$