# What is the pH when "100.0 mL" of "0.100 M HCN" is mixed with "50.0 mL" of "0.100 M NaOH"...?

Jul 15, 2018

## Class notes

see below

#### Explanation:

First ICE (in moles) table has:

• $0.0100 - 0.00500 = 0.00500$ for $\text{HCN}$
• $0.00500 - 0.00500 = 0$ for $\text{NaOH}$
• $0 + 0.00500$ for $\text{NaCN}$

Calculating for molarity, (0.005" mol")/(0.150" L")=\color(red)(0.0333)" M" of $\text{HCN}$ and/or $\text{NaCN}$ (♦)

Second ICE (in molarity) table has:

• $\setminus \textcolor{red}{0.0333} - x \ldots$ etc. for $\text{HCN}$
• $0 + x = x$ for ${\text{H"_3"O}}^{+}$
• $\setminus \textcolor{red}{0.0333} + x \ldots$ etc. for ${\text{CN}}^{-}$ (from $\setminus \textcolor{\in \mathrm{di} a n red}{\text{NaCN}}$)

(♦) I am not sure what "$\text{pOH} = - \setminus \log \left(0.0333\right) \setminus \Rightarrow 1.478$" and then "$\text{pH} = 12.5$" is, because I'm pretty sure that $0.0333 \text{ M}$ is for $\text{HCN}$ and/or $\text{NaCN}$, and not hydroxide.

Jul 15, 2018

$\text{pH} = 9.21$

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.

$0.1000 \cancel{\text{L" xx "0.100 mol"/cancel"L" = "0.0100 mol HCN}}$

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

But since "mol NaOH" = 1/2 ("mol HCN"), this yields an ideal buffer, i.e. ${\text{mol HCN" = "mol CN}}^{-}$. No ICE table is needed yet, but... here is one in mols:

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

$\text{I"" "0.01000" "" "" "0.00500" "" "" "0.00000" "" } -$
$\text{C"" "-0.00500" "-0.00500" "" "+0.00500" "" } -$
$\text{E"" "0.00500" "" "" "0.00000" "" "" "0.00500" "" } -$

Or, we could just write:

$\text{0.01000 mol HCN" - "0.00500 mol NaOH" = "0.00500 mol HCN}$ leftover

which means ${\text{0.00500 mol CN}}^{-}$ has been generated from what has been neutralized.

The current concentrations are:

$\text{0.00500 mol HCN"/("0.1000 L" + "0.0500 L") = "0.0333 M HCN}$

= "0.0333 M CN"^(-) = "0.00500 mol CN"^(-)/("0.1000 L" + "0.0500 L")

The ${K}_{a} = 6.2 \times {10}^{- 10}$ for $\text{HCN}$.

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.

${\text{HCN"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "CN"^(-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

$\text{I"" "0.0333" "" "" "" "-" "" "" "0.0333" "" "" } 0$
$\text{C"" "-x" "" "" "" "" "-" "" "" "+x" "" "" } + x$
$\text{E"" "0.0333 - x" "" "-" "" "" "0.0333+x" "" } x$

$6.2 \times {10}^{- 10} = \left(\left[\text{CN"^(-)]["H"_3"O"^(+)])/(["HCN}\right]\right)$

$= \frac{x \left(0.0333 + x\right)}{0.0333 - x}$

But since ${K}_{a} < {10}^{- 5}$, the small x approximation gives

$6.2 \times {10}^{- 10} = \frac{x \left(0.0333 + \cancel{x}\right)}{0.0333 - \cancel{x}}$

$\implies x \approx 6.2 \times {10}^{- 10} \text{M}$

Thus, with $x \equiv \left[{\text{H"_3"O}}^{+}\right]$,

$\textcolor{b l u e}{\text{pH}} = - \log \left(6.2 \times {10}^{- 10}\right) = \textcolor{b l u e}{9.21}$

which makes sense because a strong base neutralized not all of a weak acid, and thus $\text{pH} > 7$ at ${25}^{\circ} \text{C}$.