Question

what is the pH of 50.0 mL of 0.100 M HCN 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.

Answer 1

Hint: pH of any substance tells its acidic or basic nature. The ${{K}_{a}}$is the dissociation constant of an acid, that tells the extent up to which the acid dissociates. The ${{K}_{a}}$ given has error, actual ${{K}_{a}}$ for HCN is $6.17\times {{10}^{-10}}$. The cyanide anion is the conjugate base for a weak acid which is hydrogen cyanide.
Formula used: concentration = $\dfrac{no.\,of\,moles}{volume}$
Degree of dissociation, $x=\sqrt{{{K}_{b}}\times V}$ , where V is volume, ${{K}_{a}}$is the dissociation constant of an acid.
pH + pOH = 14

Complete answer:
We have been given a weak acid HCN whose concentration is 50.0 mL of 0.100 M, this when mixed with a strong base NaOH in 50.0 mL of 0.100 M, will have a pH which needs to be determined. For this we will take out the number of moles for the reaction taking place, then we will take out the dissociation constant for a base, ${{K}_{b}}$, and then through the dissociation formula the individual concentration, and hence through pH formula, the pH of the solution.
The reaction taking place is, $HCN(aq)+NaOH(aq)\to NaCN(aq)+{{H}_{2}}O(l)$
So, number of moles from the given concentration and volume will be,
Number of moles = $\dfrac{0.100\,mol\,HCN}{L}\times 0.0500L$
Number of moles = 0.00500 moles HCN = 0.00500 moles $C{{N}^{-}}$
Now, as the reaction proceeds, no base is left, so the reaction has,
\[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{N}^{-}}(aq)+{{H}_{2}}O(l)\rightleftharpoons HCN(aq)+O{{H}^{-}}(aq) \\
 & initial\,\,\,\,\,\,\,\,\,\,0.0500\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
 & final\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.0500-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \\
\end{align}\]
Now, keeping these concentrations, into the relation between ${{K}_{a}}$and ${{K}_{b}}$we have, ${{K}_{a}}\times {{K}_{b}}={{K}_{w}}$
We know that, ${{K}_{a}}$for HCN is $6.17\times {{10}^{-10}}$, ${{K}_{w}}={{10}^{-14}}$ so, rearranging for, ${{K}_{b}}$
${{K}_{b}}(C{{N}^{-}})=\dfrac{{{K}_{w}}}{{{K}_{a}}(HCN)}=\dfrac{{{10}^{-14}}}{6.2\times {{10}^{-10}}}$
${{K}_{b}}=1.61\times {{10}^{-5}}$
Now, we know that ${{K}_{b}}=\dfrac{[HCN][O{{H}^{-}}]}{[C{{N}^{-}}]}=\dfrac{{{x}^{2}}}{0.0500-x}$this value is approximately equal and in accordance with the value, ${{K}_{b}}=1.61\times {{10}^{-5}}$, so now we will take out the concentration x as, $x=\sqrt{{{K}_{b}}\times V}$ , $x=\sqrt{1.61\times {{10}^{-5}}\times 0.0500}$
As, x is the $[O{{H}^{-}}]=8.98\times {{10}^{-4}}M$ ,
Now, taking out the pOH = -log$[O{{H}^{-}}]$= 3.05
Therefore, pH = 14 – pOH
pH = 14 – 3.05
pH = 10.95
Hence, pH is 10.95

Note:
Dissociation constant, for acid, ${{K}_{a}}=\dfrac{[{{A}^{+}}][{{B}^{-}}]}{[AB]}$ , AB is the reactant, and A and B are product concentrations, while same is for the base. ${{K}_{w}}$ is the ionic product constant for water, whose value is value is ${{10}^{-14}}$ , as water consist of a pH of 7 and is neutral.