A solution P contains 0.55 mol/L ammonia and 0.10 mol/L ammonium chloride. What is the pH of the solution after the addition of 1.0 mL of 0.10 mol/L NaOH?
This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.
The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia,
A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.
This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.
Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.
The trick here is that the pH of the solution is the only information you actually need here.
Don't worry about what solution
First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide,
Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions,
So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.
#"pH" > 10.0#
Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide
#color(blue)("pH" + "pOH" = 14)#
Therefore, you have
#"pOH" = 14 - 10.0 = 4.0#
This means that the concentration of hydroxide anions is
#"pOH"= - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#
#["OH"^(-)] = 10^(-4)"M"#
Now, you take a
#color(blue)(c = n/V implies n = c * V)#
#n_(OH^(-)) = 10^(-4)"mol" color(red)(cancel(color(black)("dm"^(-3)))) * 1 color(red)(cancel(color(black)("dm"^3))) = 10^(-4)"moles OH"^(-)#
Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution
#n_(OH^(-)"added") = "0.1 mol" color(red)(cancel(color(black)("dm"^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = 10^(-4)"mol OH"^(-)#
Now, the total number of moles of hydroxide anions will be
#n_(OH^(-)"total") = n_(OH^(-)) + n_(OH^(-)"added")#
#n_(OH^(-)"total") = 10^(-4)"moles" + 10^(-4)"moles" = 2 * 10^(-4)"moles OH"^(-)#
The total volume of the resulting solution will be
#V_"total" = V_"sample P" + V_"added"#
#V_"total" = "1 dm"^3 + 1 * 10^(-3)"dm"^3 = "1.001 dm"^3#
The molarity of the hydroxide anions after you add the sodium hydroxide solution will be
#["OH"^(-)] = (2 * 10^(-4)"moles")/"1.001 dm"^3 = 1.998 * 10^(-4)"M"#
The pOH of the resulting solution will be
#"pOH" = -log(1.998 * 10^(-4)) = 3.70#
The pH of this solution will be
#"pH" = 14 - 3.70 = color(green)(10.3)#
As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.
This is a recipe for an alkaline buffer. It resists the addition of small amounts of acid and alkali keeping the pH constant.
How does it work?
You need a solution of a weak base and its salt. In this case we have ammonia solution and the salt ammonium chloride.
Ammonium chloride is an ionic salt and dissociates completely in water:
If a small amount of
In this question a small amount of
A calculation can show this:
Ammonium ions dissociate:
The number of moles of
These will combine with the
In 1 litre of P we have 0.1 mol of
So we are left with
By the same reasoning you can see that the number of moles of ammonia must have increased by this amount.
To find the pH of a buffer we rearrange
Concentration = moles/volume. We can ignore the total volume of the solution since it is common to both
So the pH does not change.
This is for illustration only. As you state, this is a multi - choice question so my take on this is that you are required recognise it is the recipe for an alkali buffer so reason that the pH won't change.