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:
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 that0.0333" 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