Question

A 1.00L buffer solution is 0.150M in $H{{C}_{7}}{{H}_{5}}{{O}_{2}}$ and 0.250M in $Li{{C}_{7}}{{H}_{5}}{{O}_{2}}$. How do you calculate the pH of the solution after the addition of 100.0mL of 1.00M $HCl$ ? The ${{K}_{a}}$ for $H{{C}_{7}}{{H}_{5}}{{O}_{2}}$is $6.5\times {{10}^{-5}}$.

Answer 1

Hint: The answer here is based on the simple calculation of pH of the solution where pH of the solution which is defined as the negative logarithm of the hydrogen ion concentration. Calculate $p{{K}_{a}}$ followed by calculating pH.

Complete answer:
In the previous classes, we have studied about the concepts of physical chemistry which tells about several simple calculations like calculation of molality, molarity, normality, pH of a solution and many other related chapters.
Let us see some of these concepts in detail and deduce the required answer.
- pH of a solution is defined as the negative logarithm of hydrogen ion concentration. If the value in pH scale is below 7 then the solution is acidic and if the pH of a solution is more than 7 then that solution is basic in nature.
- Now by looking into the question we can say that the given compound is benzoic acid where the formula $H{{C}_{7}}{{H}_{5}}{{O}_{2}}$ can be written further as ${{C}_{6}}{{H}_{5}}COOH$
- Now, this benzoic acid dissociates as,
\[{{C}_{6}}{{H}_{5}}COOH{{C}_{6}}{{H}_{5}}CO{{O}^{-}}+{{H}^{+}}\] ................(1)
Since, the formula $Li{{C}_{7}}{{H}_{5}}{{O}_{2}}$ can be written as,${{C}_{6}}{{H}_{5}}COOLi$, the lithium ion provides the large reserve of a base by the complete dissociation. This is written as,
\[{{C}_{6}}{{H}_{5}}COOLi{{C}_{6}}{{H}_{5}}CO{{O}^{-}}+L{{i}^{+}}\] ................(2)
Let us consider ${{C}_{6}}{{H}_{5}}CO{{O}^{-}}$ as${{A}^{-}}$ . Now, initial moles of ${{A}^{-}}$is given by,
$nA_{^{\operatorname{in}it}}^{-}=1\times 0.25=0.25$
Now, number of${{H}^{+}}$ added is given by,
$n{{H}^{+}}=1\times \dfrac{100}{1000}=0.1$
From equation (1) we can see that hydrogen ions will be absorbed by a large reserve of${{A}^{-}}$ions which are shifting the equilibrium to the left and therefore buffering pH.
Here, they react in 1 : 1 molar ratio so that number of moles of ${{A}^{-}}$remaining will be,
$n{{A}^{-}}=0.25-0.1=0.15$
Thus each mole of hydrogen ion added will form one mole of $HA$ and hence the number of moles of $HA$ formed is 0.1.
Now, initial moles of $HA$is given by,
$nH{{A}_{init}}=0.15\times 1=0.15$
Thus, the total moles of $HA$is given by,
$nHA-0.1+0.15=0.25$
We know that acid dissociation constant is given by,
\[{{K}_{a}}=\dfrac{\left[ H_{aq}^{+} \right]\left[ A_{aq}^{-} \right]}{\left[ H{{A}_{aq}} \right]}\]
By rearranging this equation, we get
$\left[ H_{aq}^{+} \right]={{K}_{a}}\dfrac{\left[ HA \right]}{\left[ A_{aq}^{-} \right]}$
We are given the data as acid dissociation constant is$6.5\times {{10}^{-5}}$
Now, substituting these values in above equation we get
$\left[ H_{aq}^{+} \right]=6.5\times {{10}^{-5}}\times \dfrac{\left[ 0.25 \right]}{\left[ 0.15 \right]}=1.083\times {{10}^{-4}}$
Now, we know that $pH=-\log \left[ {{H}^{+}} \right]=-\log \left[ 1.083\times {{10}^{-4}} \right]$
This will give, pH = 3.96

Note:
Note the difference in pH and $p{{K}_{a}}$ lies in fact that pH is the measure of concentration of hydrogen ions in aqueous solution whereas $p{{K}_{a}}$is acid dissociation constant which helps you to predict what a molecule will do at specific pH. This relationship is given by the Henderson-Hasselbalch equation.