Question

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

`"NaOH"`, `"HBrO"`, `"NH"_3`, `"Sr"("OH")_2`, `"HBr"`

Answer 1

`"Sr"("OH")_2(aq) > "NaOH"(aq) > "NH"_3(aq) > "HBrO"(aq) > "HBr"(aq)`

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Here's what we should know:

• The stronger the base, the higher the `"pH"`, since it dissociates more in solution than weaker bases, thus making it more basic and raising the `"pH"`.

• The stronger the acid, the lower the `"pH"`, since it dissociates more in solution than weaker acids, thus making it more acidic and lowering the `"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`
• `"NaOH"`

The weak base within this list is `"NH"_3`.
The strong acid within this list is `"HBr"`.
The weak acid within this list is `"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 `"OH"^(-)` as `"NaOH"` is.

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

`"Sr"("OH")_2(aq) > "NaOH"(aq) > ? > ? > "HBr"(aq)`

We now know that `"HBrO"` is a weak acid, so it must dissociate less than `"HBr"`, meaning that it decreases the `"pH"` by less from about `7`. That means the `"pH"` of `"HBrO"(aq) > "pH"` of `"HBr"(aq)`.

Now we have:

`"Sr"("OH")_2(aq) > "NaOH"(aq) > ? > "HBrO"(aq) > "HBr"(aq)`

Since `"NH"_3` is a weak base, it increases the `"pH"` by less than `"NaOH"`, but since it increases the `"pH"` and `"HBrO"` decreases the `"pH"`, it means that in general the same concentration of both implies that the `"pH"` is higher for `"NH"_3(aq)` than `"HBrO"(aq)`.

Therefore, our result is:

`bb("Sr"("OH")_2(aq) > "NaOH"(aq) > "NH"_3(aq) > "HBrO"(aq) > "HBr"(aq))`