Question

Which of the following aqueous solutions has the lowest `[OH^-]` ?

A. a solution with a pH of 3.0
B. a `1*10^-4` M solution of `HNO_3`
C. a solution with a pOH of 12.0
D. pure water
E. a `1*10^-6`M solution of NaOH

Answer 1

Would it not be `C.`?

Explanation:

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

`2H_2O(l) rightleftharpoons H_3O^+ + HO^-`

And `K_w=[H_3O^+][HO^-]=10^-14`...and if we take `log_10` of both sides..we get...

`14=pH+pOH`

And so we gots... `A. [H_3O^+]=0.001*mol*L^-1`

`B. [H_3O^+]=0.0001*mol*L^-1`

`C. pOH=12,`...so `pH=2`, and thus `[H_3O^+]=10^-2*mol*L^-1-=0.01*mol*L^-1`

`D.` `pH=7` and thus `[H_3O^+]=10^-7*mol*L^-1`

`E.` `[HO^-]=10^-6*mol*L^-1`, `pOH=+6`, `pH=8`, and so `[H_3O^+]=10^-8*mol*L^-1`

And so the HIGHEST concentration with respect to `H_3O^+` NECESSARILY demands the LOWEST CONCENTRATION with respect to `HO^-`. Note that we use `H_3O^+` and `H^+` interchangeably...

Capisce?

Answer 2

My guess is `C.`

Explanation:

Let's examine the options by looking at their `"pH"`.

`A.` has a `"pH"=3`.

For `B.`, the `"pH"` of the nitric acid would be `-log[10^-4]=4`, as it is an acid, and so the equation corresponds to the `"pH"`.

For `C.`, the `"pH"` of the solution would be `14-"pOH"=14-12=2`.

`D.` is pure water, and therefore has a `"pH"=7`.

For `E.` the `"pOH"` of the sodium hydroxide would be `-log[10^-6]=6`, as it is a base, and so the logarithm equation responds to the `"pOH"`.

Since the `"pH"` is related to `"pOH"` by the equation,

`"pH + pOH"=14` at `25^@"C"`, we can find the `"pOH"` of the following choices.

`A.` has a `"pOH"=14-3=11`.

`B.` has a `"pOH"=14-4=10`.

`C.` has a `"pOH"=12`, as stated.

`D.` has a `"pOH"=14-7=7`.

`E.` has a `"pOH"=6`, as calculated.

Notice how,

`6<7<10<11<12`

or

`E.<D.<B.<A.<C.`

Since `"pOH"=-log[OH^-]`, or `[OH^-]=10^(-"pOH")`, and so the hydroxide ion concentration is inversely proportional to the `"pOH"`, i.e. more `[OH^-]` means less `"pOH"`, and vice-versa.

Since `C.` has the highest `"pOH"`, then that means that it has the least `[OH^-]` concentration, and so it is the answer.