Question

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

Answer 1

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...?

Answer 2

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

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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)`