What is the pH if 50.0 mL of 0.100 M HCN is mixed with 50.0 mL of 0.100 M NaOH?

Given: #K_a# for HCN is #1.00xx10^-6#
which is probably in error

2 Answers
Jul 3, 2018

Answer:

And so at the end of the reaction we have #0.100*mol*L^-1# #K^+""^(-)C-=N(aq)# in #100*mL# of solution... I get #pH=8.90#

Explanation:

And cyanide anion, the conjugate base of a WEAK acid will cause water hydrolysis to give equilibrium quantities of #H-C-=N#, and #KOH(aq)#...i.e we address the equation...

#H_2O(l) + ""^(-)C-=N rightleftharpoonsHC-=N(aq) + HO^-#

This site reports that #K_a(HCN)=6.17xx10^-10#

And so if the degree of association is #x#...then...

#K_a=([HC-=N][HO^-])/([""^(-)C-=N])=6.17xx10^-10#

#6.17xx10^-10-=x^2/(0.100-x)#...and of #x# is small...then #0.100-x~=0.100#

And so ………………

#x_1=sqrt(6.17xx10^-10xx0.100)=7.85xx10^-6*mol*L^-1#..

#x_2=sqrt(6.17xx10^-10xx(0.100-7.85xx10^-6))=7.85xx10^-6*mol*L^-1#..

And so near enuff is good enuff. But #x=[HO^-]#...#pOH=-log_10(7.85xx10^-6)=5.11#...#pH=14-pOH=8.90#.

#pH# is slightly elevated from #7# given that cyanide is an (admittedly weak) base....do you follow...?

Jul 3, 2018

Here is an alternative approach and explanation to anor's answer, but I get #"pH" = 10.95#.


When #"HCN"# and #"NaOH"# react, the #"HCN"# is neutralized exactly, since both concentrations and volumes are identical for a monoprotic acid reacting with a base containing one #"OH"^(-)#:

#"HCN"(aq) + "NaOH"(aq) -> "NaCN"(aq) + "H"_2"O"(l)#

First, it is important to note that a dilution has occurred upon full reaction!

It should be noted that

#"0.100 mol HCN"/cancel"L" xx 0.0500 cancel"L" = "0.00500 mols HCN"#

#harr "0.00500 mols CN"^(-)#

is contained in the #"100. mL"# total volume (#"0.100 L"#), which gives a #ul("0.0500 M CN"^(-))# solution (represented as #"NaCN"#).

Since no other base is left in solution (all the #"NaOH"# is gone), we proceed to the association of #"CN"^(-)# in water as the main reaction of interest at this point, which gives us our ICE table:

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

#"I"" "0.0500" "" "" "-" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "" "-" "" "+x" "" "" "+x#
#"E"" "0.0500-x" "-" "" "" "x" "" "" "" "x#

This requires

  • that we know #K_a# for #"HCN"# is #6.2 xx 10^(-10)#.
  • that we know #K_b#, for #"CN"^(-)# is a base and is in water now.

At #25^@ "C"#, we can say that

#K_b ("CN"^(-)) = K_w/(K_a("HCN")) = 10^(-14)/(6.2 xx 10^(-10)) = 1.61 xx 10^(-5)#

so that

#1.61 xx 10^(-5) = (["HCN"]["OH"^(-)])/(["CN"^(-)])#

#= (x^2)/(0.0500 - x)#

Making the small #x# approximation is good here, as #K_b# is on the order of #10^(-5)# (even though #["CN"^(-)]_i# is somewhat small!). So:

#1.61 xx 10^(-5) ~~ x^2/0.0500#

#=> x = sqrt(0.0500 cdot 1.61 xx 10^(-5))#

#= ["OH"^(-)] = 8.98 xx 10^(-4) "M"#

Therefore, from here we can find the #"pOH"# to be:

#"pOH" = -log["OH"^(-)] = 3.05#,

and at this temperature,

#color(blue)("pH") = 14 - "pOH" = color(blue)(10.95)#