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

Class notes

see below

Explanation:

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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".