Question

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?)

Answer 1

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.

Answer 2

`"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"`.