Find the #"pH"# in each of the following scenarios? #"pK"_a = 3.36# for #"HNO"_2#.

DISCLAIMER: REALLY LONG ANSWER!

I got #"pH" = 3.65, 3.35, 4.05, 8.42, 1.88, 12.48#.

#a)#

Well, using their molar masses:

#"4.70 g HNO"_2 xx ("1 mol")/("47.0129 g") ~~ "0.100 mols"#

#"13.8 g NaNO"_2 xx ("1 mol")/("70.0019 g") ~~ "0.197 mols"#

It's all in #"1 L"#, but we don't have to care because in a buffer, the volume in their ratio of concentrations cancel out.

The Henderson-Hasselbalch equation gives for a #"pK"_a# of #3.39# for #"HNO"_2#. I'll use #3.36# since it gets me right onto your answer key, and it's not 'fudging' since it's within reason.

#color(blue)("pH") = "pK"_a + log((["base"])/(["acid"]))#

#= 3.36 + log((0.197)/(0.100))#

#= color(blue)(3.65)#

#b)# Now #"100 mL"# of buffer are present. That decreases the number of mols we have by a factor of #10#. We assume that the total volume doesn't change, so that we only look at the change in mols.

• Adding strong acid neutralizes the molar equivalent of the weak base and generates that many mols of weak conjugate acid.

• Adding strong base neutralizes the molar equivalent of the weak acid and generates that many mols of weak conjugate base.

#(i)#

#color(blue)("pH") = 3.36 + log(("0.0197 mols NO"_2^(-) - "0.00500 mols HCl")/("0.0100 mols HNO"_2 + "0.00500 mols HCl"))#

#= color(blue)(3.35)#

As we can now see, the smaller number of mols decreases the buffer capacity (the tolerance to incoming strong acid/base).

#(ii)#

It's basically the opposite for adding #"5.00 mmol NaOH"#. The rationale is at the top of #(b)#, and of note is that the #+//-# switch places.

#color(blue)("pH") = 3.36 + log(("0.0197 mols NO"_2^(-) + "0.00500 mols NaOH")/("0.0100 mols HNO"_2 - "0.00500 mols NaOH"))#

#= color(blue)(4.05)#

#(iii)# Almost the same thing as #(ii)#, but with twice as much #"NaOH"#.

#"400 mg NaOH" xx ("1 mol")/"39.9959 g NaOH"#

#= "10.00 mmol" = "0.01000 mols NaOH"#

The problem is, now all the #"HNO"_2# is neutralized, so we now have:

#"0.0197 mols NO"_2^(-) + "0.01000 mols neutralized HNO"_2#

#= "0.0297 mols NO"_2^(-)#

This #"NO"_2^(-)# now reacts backwards in an association equilibrium. Using the #"pK"_a# of #"HNO"_2#, at #25^@ "C"# and #"1 atm"#,

#"pK"_b = "pK"_w - "pK"_a = 14 - 3.36 = 10.64#

And so, #K_b = 10^(-"pK"_b) = 2.29 xx 10^(-11)#. The dissociation is typical in water and should become routine. The base grabs a proton and generates #"OH"^(-)# in water.

The ICE table in #"mols"# is:

#"NO"_2^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HNO"_2(aq) + "OH"^(-)(aq)#

#"I"" "0.0297" "" "" "-" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "" "-" "" "+x" "" "" "+x#
#"E"" "0.0297-x" "-" "" "" "x" "" "" "" "x#

And so,

#2.29 xx 10^(-11) = (["HNO"_2]["OH"^(-)])/(["NO"_2^(-)])#

This #K_b# is so small, that #x# #"<<"# #"0.0297 mols"#. Thus, and remember that #K_b# takes concentrations,

#2.29 xx 10^(-11) ~~ ((x/"0.1000 L buffer")(x/"0.1000 L buffer"))/(("0.0297 mols"/("0.1000 L buffer")))#

#= (x^2)/("0.1000 L")^2*"0.1000 L"/"0.0297 mols"#

#= x^2/("0.00297 mol"cdot"L")#

And

#x ~~ sqrt(2.29 xx 10^(-11) "mol/L" cdot "0.00297 mol"cdot"L")#

#= 2.61 xx 10^(-7) "mols"#.

As a result, the new #"pH"# is based on #"OH"^(-)#:

#color(blue)("pH") = 14 - "pOH" = 14 + log["OH"^(-)]#

#= 14 + log((2.61 xx 10^(-7) "mols")/"0.1000 L buffer")#

#= color(blue)(8.42) ~~ 8.41#

#(iv)# Now we instead use #"730 mg"# of #"HCl"#. Lemme guess... it neutralizes #"NO"_2^(-)# completely.

#"730 mg HCl" xx ("1 mol")/"36.4609 g" = "20.02 mmol" = "0.02002 mmol HCl"#

Yep, we had #"0.0197 mols NO"_2^(-)#, so we now lose all of it to make #"0.0197 mols HNO"_2#, and also have #3.21 xx 10^(-4) "mols H"^(+)# remaining as the initial concentration in the acid dissociation.

The ICE table in #"mols"# is:

#"HNO"_2(aq) + "H"_2"O"(l) rightleftharpoons "NO"_2^(-)(aq) + "H"^(+)(aq)#

#"I"" "0.0297" "" "" "-" "" "" "0" "" "" "" "3.21 xx 10^(-4)#
#"C"" "-x" "" "" "" "-" "" "+x" "" "" "+x#
#"E"" "0.0297-x" "-" "" "" "x" "" "" "" "3.21 xx 10^(-4) + x#

Now, the #K_a# was #10^(-"pK"_a) = 10^(-3.36) = 4.37 xx 10^(-4)#, so:

#4.37 xx 10^(-4) = (x/"0.1000 L"(3.21 xx 10^(-4) + x)/"0.1000 L")/((0.0297 - x)/"0.1000 L")#

#= (x(3.21 xx 10^(-4) + x))/((0.0297 - x)cdot0.1000)#

Base on the size of #K_a# vs. #["H"^(+)]_i#, we approximate that #x# is small compared to #0.0297#, but not compared to #3.21 xx 10^(-4)#. This gives:

#x^2 + 3.21 xx 10^(-4)x - 4.37 xx 10^(-4) cdot 0.00297 ~~ 0#

Solving this quadratic gives #x = "0.00099 mols"# (compared to the true value of #"0.00097 mols"#). Apparently it was a great approximation!

So now,

#["H"^(+)] ~~ (3.21 xx 10^(-4) "mols" + "0.00099 mols")/("0.1000 L") = "0.01311 M H"^(+)#

And this #"pH"# is:

#color(blue)("pH") = -log["H"^(+)] = color(blue)(1.88)#

I think this is more accurate of an answer than what is given. The difference is precisely due to the choice from the answer key of ignoring #["H"^(+)]_i#.

However, I don't think that is sensible because you can see that the #K_a# is on the order of #10^(-4)#, and so is #["H"^(+)]_i#. So, I think the answer key is wrong in this case.

[If #["H"^(+)]_i = "0.00 M"#, then I get #"pH" ~~ 1.94#.]

#(v)# You can infer off of problem #(iii)# that there is an extra #"120 mg NaOH"#. Therefore, we go past the equivalent point just like in #(iii)#, by this amount:

#"120 mg NaOH" xx "1 mol"/"39.9959 g" = "3.00 mmols"#

#=# #"0.00300 mols NaOH"# extra

Now we construct yet another ICE table in #"mols"#! We neutralized all the #"HNO"_2# and make that many more mols of #"NO"_2^(-)#, and have #"0.00300 mols NaOH"# extra.

#"NO"_2^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HNO"_2(aq) + "OH"^(-)(aq)#

#"I"" "0.0297" "" "" "-" "" "" "0" "" "" "" "0.00300#
#"C"" "-x" "" "" "" "-" "" "+x" "" "" "+x#
#"E"" "0.0297-x" "-" "" "" "x" "" "" "" "0.00300 + x#

#K_b = 2.29 xx 10^(-11) = (["HNO"_2]["OH"^(-)])/(["NO"_2^(-)])#

#= ((x(0.00300 + x))/("0.1000 L"cdot "0.1000 L"))/((0.0297 - x)/("0.1000 L"))#

#= (x(0.00300 + x))/((0.0297 - x)cdot0.1000)#

As usual, we now assume that #x# #"<<"# #0.0297# and #0.00300# since #K_b# is so puny.

#K_b ~~ (x(0.00300))/((0.0297)cdot0.1000)#

#=> x ~~ 2.27 xx 10^(-11) "mols"#

And so, #["OH"^(-)] = ("0.00300 mols" + 2.27 xx 10^(-11) "mols")/"0.1000 L"# and

#=>["OH"^(-)] ~~ "0.0300 M"#

This gives

#"pOH" = -log["OH"^(-)] = 1.52#

And so,

#color(blue)("pH") = 14 - "pOH" = 14 - 1.52#

#= color(blue)(12.48)#

#c)# I got values of #"pH" = 3.65, 3.35, 4.05, 8.42, 1.88, 12.48#.

• The first three are cases where the buffer is not demolished.
• The fourth one is where the buffer was neutralized of all the weak acid, and so it should be base-ic #"pH"#.
• The fifth one is where the buffer was neutralized of all the weak base, and so it should be acidic #"pH"#.
• The sixth one is where the buffer was neutralized of all the weak acid, but even more base was added to get pretty far past the equivalence point, so this makes sense to be really base-ic #"pH"#.