What is the pH when #"100.0 mL"# of #"0.100 M HCN"# is mixed with #"50.0 mL"# of #"0.100 M NaOH"#...?

"...if the #K_a# for #HCN# is #1.00xx10^-6#?"

(Ignoring the mistaken constant part, how would I set up the ICE table(s) for this?)

2 Answers
Jul 15, 2018

Answer:

Class notes

see below

Explanation:

webcam

First ICE (in moles) table has:

  • #0.0100-0.00500=0.00500# for #"HCN"#
  • #0.00500-0.00500=0# for #"NaOH"#
  • #0+0.00500# for #"NaCN"#

    Calculating for molarity, #(0.005" mol")/(0.150" L")=\color(red)(0.0333)" M"# of #"HCN"# and/or #"NaCN"# (♦)

Second ICE (in molarity) table has:

  • #\color(red)(0.0333)-x...# etc. for #"HCN"#
  • #0+x=x# for #"H"_3"O"^(+)#
  • #\color(red)(0.0333)+x...# etc. for #"CN"^(-)# (from #\color(indianred)("NaCN")#)

    (♦) I am not sure what "#"pOH"=-\log(0.0333)\rArr1.478#" and then "#"pH"=12.5#" is, because I'm pretty sure that #0.0333" M"# is for #"HCN"# and/or #"NaCN"#, and not hydroxide.

Jul 15, 2018

#"pH" = 9.21#


Well, since the volume of strong base is less than the volume of weak acid AND both are the same concentration, this forms a buffer.

#0.1000 cancel"L" xx "0.100 mol"/cancel"L" = "0.0100 mol HCN"#

#0.0500 cancel"L" xx "0.100 mol"/cancel"L" = "0.00500 mol NaOH"#

But since #"mol NaOH" = 1/2 ("mol HCN")#, this yields an ideal buffer, i.e. #"mol HCN" = "mol CN"^(-)#. No ICE table is needed yet, but... here is one in mols:

#"HCN"(aq) " "+" " "OH"^(-)(aq) rightleftharpoons "CN"^(-)(aq) + "H"_2"O"(l)#

#"I"" "0.01000" "" "" "0.00500" "" "" "0.00000" "" "-#
#"C"" "-0.00500" "-0.00500" "" "+0.00500" "" "-#
#"E"" "0.00500" "" "" "0.00000" "" "" "0.00500" "" "-#

Or, we could just write:

#"0.01000 mol HCN" - "0.00500 mol NaOH" = "0.00500 mol HCN"# leftover

which means #"0.00500 mol CN"^(-)# has been generated from what has been neutralized.

The current concentrations are:

#"0.00500 mol HCN"/("0.1000 L" + "0.0500 L") = "0.0333 M HCN"#

#= "0.0333 M CN"^(-) = "0.00500 mol CN"^(-)/("0.1000 L" + "0.0500 L")#

The #K_a = 6.2 xx 10^(-10)# for #"HCN"#.

Write the expression for the acid dissociation since there is weak acid still present and it now dissociates after being halfway neutralized. This time, this is in molarity.

#"HCN"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "CN"^(-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" "0.0333" "" "" "" "-" "" "" "0.0333" "" "" "0#
#"C"" "-x" "" "" "" "" "-" "" "" "+x" "" "" "+x#
#"E"" "0.0333 - x" "" "-" "" "" "0.0333+x" "" "x#

#6.2 xx 10^(-10) = (["CN"^(-)]["H"_3"O"^(+)])/(["HCN"])#

#= (x(0.0333+x))/(0.0333-x)#

But since #K_a < 10^(-5)#, the small x approximation gives

#6.2 xx 10^(-10) = (x(0.0333+cancelx))/(0.0333-cancelx)#

#=> x ~~ 6.2 xx 10^(-10) "M"#

Thus, with #x -= ["H"_3"O"^(+)]#,

#color(blue)("pH") = -log(6.2 xx 10^(-10)) = color(blue)(9.21)#

which makes sense because a strong base neutralized not all of a weak acid, and thus #"pH" > 7# at #25^@ "C"#.