Question #bf38f
1 Answer
Explanation:
For starters, you know that your buffer contains ammonia,
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
Now, you're working with a
This means that the initial solution contained
When you add the sodium hydroxide, the hydroxide anions will react with the ammonium cations in a
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
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" = "p"K_b + log(( ["NH"_4^(+)])/(["NH"_3]))#
This would make the
#"pH" = 14 - ["p"K_b + log(( ["NH"_4^(+)])/(["NH"_3]))]#
The
#"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" = 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"_ "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#