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?)
"...if the
(Ignoring the mistaken constant part, how would I set up the ICE table(s) for this?)
2 Answers
Class notes
see below
Explanation:
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.
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
#"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"# leftoverwhich 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
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
#6.2 xx 10^(-10) = (x(0.0333+cancelx))/(0.0333-cancelx)#
#=> x ~~ 6.2 xx 10^(-10) "M"#
Thus, with
#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