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

## Given: ${K}_{a}$ for HCN is $1.00 \times {10}^{-} 6$ I'm asked to work with that value Attempting first ICE table: $H C N + N a O H \setminus \to {H}_{2} O + N a C N$ (I) $0.0050 \text{ " 0.0100 " " " || " " } 0$ (C)$- 0.0100 \text{ -0.0100 " " || " " } + 0.0100$ (E)$- 0.0050 \text{ " " " 0 " " " || " " } 0.0100$

Jul 5, 2018

$\text{pH} = 12.523$

#### Explanation:

${K}_{b} \equiv {K}_{w} / {K}_{a} = 1.00 \times {10}^{- 8}$

The strong base $\text{NaOH}$ is in excess and neutralizes all $\text{HA}$ initially present in the solution. As a result, the hydrolysis of ${\text{A}}^{-}$, the product of the neutralization reaction. would be the only reversible process capable of generating an equilibrium.

"A"^(-)(aq) + "H"_2"O"(aq)\ color(purple)(rightleftharpoons) \ "HA"(aq) + ul("OH"^(-) (aq))

R "A"^(-)(aq) + "H"_2"O"(aq)\ color(purple)(rightleftharpoons) \ "HA"(aq) + ul("OH"^(-) (aq))
I $0.005 \textcolor{w h i t e}{- - - - - - - - - - - -} 0.005$
C $- x \textcolor{w h i t e}{- - - - - - - l l} + x \textcolor{w h i t e}{- - - -} + x$
E $0.005 - x \textcolor{w h i t e}{- - - - - l -} x \textcolor{w h i t e}{- - - l l} 0.005 + x$

(in $m o l$)

$c = \frac{n}{\textcolor{p u r p \le}{V}}$ and therefore

K_b("A"^(-)) = (overbrace(["HA"])^("conjugate acid") * ["OH"^(-)])/(overbrace(["A"^(-)])^("weak base"))\ = ((x)/(0.150)*(0.005+x))/(0.005 -x )

$\frac{x \cdot \left(0.005 + x\right)}{0.005 - x} = 1.00 \times {10}^{- 8} \cdot 0.150$

${x}^{2} + \left(0.005 + 1.00 \times {10}^{- 8}\right) \cdot x - 0.005 \cdot 1.00 \times {10}^{- 8} = 0$

$x = 1.50 \times {10}^{- 9} \textcolor{w h i t e}{l} m o l$

($x > 0$ given that $\left[\text{HA}\right] \ge 0$)

Therefore

$n \left({\text{OH}}^{-}\right) = 0.005 + 1.50 \cdot {10}^{- 9} \approx 0.005 \textcolor{w h i t e}{l} m o l$
$c \left({\text{OH}}^{-}\right) = \frac{n}{V} = 0.0333 \textcolor{w h i t e}{l} m o l \cdot {\mathrm{dm}}^{- 3}$

$\text{pH" = "pKw" - "pOH} = 12.523$

Note that ${\text{OH}}^{-}$, already present at great quantity, is the only factor through which the equilibrium influences the $\text{pH}$ of the solution. ${K}_{b} = 1.00 \times {10}^{- 8}$; thus there's a small deviation in $\text{pH}$ from what would have been observed with a strong acid of the same volume and concentration.