# 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 $1.00 \times {10}^{-} 6$ which is probably in error

Jul 3, 2018

And so at the end of the reaction we have $0.100 \cdot m o l \cdot {L}^{-} 1$ K^+""^(-)C-=N(aq) in $100 \cdot m L$ of solution... I get $p H = 8.90$

#### Explanation:

And cyanide anion, the conjugate base of a WEAK acid will cause water hydrolysis to give equilibrium quantities of $H - C \equiv N$, and $K O H \left(a q\right)$...i.e we address the equation...

H_2O(l) + ""^(-)C-=N rightleftharpoonsHC-=N(aq) + HO^-

This site reports that ${K}_{a} \left(H C N\right) = 6.17 \times {10}^{-} 10$

And so if the degree of association is $x$...then...

K_a=([HC-=N][HO^-])/([""^(-)C-=N])=6.17xx10^-10

$6.17 \times {10}^{-} 10 \equiv {x}^{2} / \left(0.100 - x\right)$...and of $x$ is small...then $0.100 - x \cong 0.100$

And so ………………

${x}_{1} = \sqrt{6.17 \times {10}^{-} 10 \times 0.100} = 7.85 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$..

${x}_{2} = \sqrt{6.17 \times {10}^{-} 10 \times \left(0.100 - 7.85 \times {10}^{-} 6\right)} = 7.85 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$..

And so near enuff is good enuff. But $x = \left[H {O}^{-}\right]$...$p O H = - {\log}_{10} \left(7.85 \times {10}^{-} 6\right) = 5.11$...$p H = 14 - p O H = 8.90$.

$p H$ is slightly elevated from $7$ given that cyanide is an (admittedly weak) base....do you follow...?

Jul 3, 2018

Here is an alternative approach and explanation to anor's answer, but I get $\text{pH} = 10.95$.

When $\text{HCN}$ and $\text{NaOH}$ react, the $\text{HCN}$ is neutralized exactly, since both concentrations and volumes are identical for a monoprotic acid reacting with a base containing one ${\text{OH}}^{-}$:

$\text{HCN"(aq) + "NaOH"(aq) -> "NaCN"(aq) + "H"_2"O} \left(l\right)$

First, it is important to note that a dilution has occurred upon full reaction!

It should be noted that

$\text{0.100 mol HCN"/cancel"L" xx 0.0500 cancel"L" = "0.00500 mols HCN}$

$\leftrightarrow {\text{0.00500 mols CN}}^{-}$

is contained in the $\text{100. mL}$ total volume ($\text{0.100 L}$), which gives a $\underline{{\text{0.0500 M CN}}^{-}}$ solution (represented as $\text{NaCN}$).

Since no other base is left in solution (all the $\text{NaOH}$ is gone), we proceed to the association of ${\text{CN}}^{-}$ in water as the main reaction of interest at this point, which gives us our ICE table:

${\text{CN"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HCN"(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" "0.0500" "" "" "-" "" "" "0" "" "" "" } 0$
$\text{C"" "-x" "" "" "" "-" "" "+x" "" "" } + x$
$\text{E"" "0.0500-x" "-" "" "" "x" "" "" "" } x$

This requires

• that we know ${K}_{a}$ for $\text{HCN}$ is $6.2 \times {10}^{- 10}$.
• that we know ${K}_{b}$, for ${\text{CN}}^{-}$ is a base and is in water now.

At ${25}^{\circ} \text{C}$, 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

$1.61 \times {10}^{- 5} = \left(\left[{\text{HCN"]["OH"^(-)])/(["CN}}^{-}\right]\right)$

$= \frac{{x}^{2}}{0.0500 - x}$

Making the small $x$ approximation is good here, as ${K}_{b}$ is on the order of ${10}^{- 5}$ (even though ${\left[{\text{CN}}^{-}\right]}_{i}$ is somewhat small!). So:

$1.61 \times {10}^{- 5} \approx {x}^{2} / 0.0500$

$\implies x = \sqrt{0.0500 \cdot 1.61 \times {10}^{- 5}}$

= ["OH"^(-)] = 8.98 xx 10^(-4) "M"

Therefore, from here we can find the $\text{pOH}$ to be:

"pOH" = -log["OH"^(-)] = 3.05,

and at this temperature,

color(blue)("pH") = 14 - "pOH" = color(blue)(10.95)