Question

Question #bf38f

Answer 1

`"pH" = 9.31`

Explanation:

For starters, you know that your buffer contains ammonia, `"NH"_3`, a weak base, and the ammonium cation, `"NH"_4^(+)`, its conjugate acid.

When you add sodium hydroxide, a strong base, the hydroxide anions delivered to the solution by the sodium hydroxide will react with the ammonium cations to form ammonia and water

`"NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-) -> "NH"_ (3(aq)) + "H"_ 2"O"_ ((l))`

As you can see, the buffer will convert a strong base to a weak base, which is why you should expect the to see a small increase, much smaller than what you see when sodium hydroxide is added to pure water, in the `"pH"` of the solution.

Now, you're working with a `"1.00-L"` buffer, so you can treat molarity and number of moles interchangeably.

This means that the initial solution contained `0.15` moles `"NH"_3` and `0.15` moles `"NH"_4^(+)`.

When you add the sodium hydroxide, the hydroxide anions will react with the ammonium cations in a `1:1` mole ratio. The reaction will also produce ammonia in a `1:1` mole ratio.

This means that after the reaction takes place, the buffer will contain

`n_ ("OH"^(-)) = "0 moles " ->` completely consumed

`n_ ("NH"_ 4^(+)) = "0.15 moles" - "0.010 moles" = "0.014 moles NH"_4^(+)`

`n_ ("NH"_ 3) = "0.15 moles" + "0.010 moles" = "0.16 moles NH"_3`

In other words, every mole of hydroxide anions added to the buffer will consume `1` mole of ammonium cations and produce `1` mole of ammonia.

If you assume that the volume of the buffer does not change, you will end up with

`["NH"_3] = "0.16 M" " "` and `" " ["NH"_4^(+)] = "0.14 M"`

Now, you can use the Henderson-Hasselbalch equation to calculate the `"pOH"` of the buffer

`"pOH" = "p"K_b + log(( ["NH"_4^(+)])/(["NH"_3]))`

This would make the `"pH"` of the solution equal to

`"pH" = 14 - ["p"K_b + log(( ["NH"_4^(+)])/(["NH"_3]))]`

The `"p"K_b` of the weak base is defined as

`"p"K_b = -log(K_b)`

In your case, this will be equal to

`"p"K_b = - log(1.76 * 10^(-5)) = 4.75`

Therefore, you can say that the `"pH"` of the solution after the addition of the strong base is equal to

`"pH" = 14 - [4.75 + log((0.14 color(red)(cancel(color(black)("M"))))/(0.16color(red)(cancel(color(black)("M")))))] = color(darkgreen)(ul(color(black)(9.31)))`

The answer is rounded to two decimal places, the number of sig figs you have for your values.

Notice that the `"pH"` of the solution before adding the strong base was lower than `9.31`

`"pH"_ "before adding NaOH" = 14 - [4.75 + log ( color(red)(cancel(color(black)("0.15 M")))/color(red)(cancel(color(black)("0.15 M"))))]`

`"pH"_ "before adding NaOH" = 14 - 4.75 = 9.25`