Question

How would you use the Henderson-Hasselbalch equation to calculate the $pH$ of the solution? A solution that is $0.800\% {\text{ }}{{\text{C}}_5}{H_5}N$ by mass and $0.990\% {\text{ }}{{\text{C}}_5}{H_5}NHCl$ by mass.

Answer 1

Hint: We must first know what an Henderson-Hasselbalch equation is. The Henderson-Hasselbalch equation is helpful in estimating the pH of a buffer solution and helps in finding the equilibrium $pH$ in an acid base reaction. This equation can be used to determine the amount of an acid and the conjugate base which is needed to make a buffer solution of a certain $pH$. $pKa$is defined as the quantitative measure of the strength of an acid in solution.

Complete step-by-step answer:The Henderson-Hasselbalch equation connects the$pH,pKa$ and the molar concentration together.
This equation helps in finding out the $pH$of the buffer solution. The equilibrium which exists between the weak acid and the conjugate base allows the solution to resist the $pH$ change even when a small amount of strong acid or base is added and hence the buffer $pH$ can be estimated from,
$pH = pKa + {\log _{10}}\left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right)$
Where, $pKa$ is the negative algorithm of $Ka$
$Ka$ is the acid dissociation constant
$HA$ is the concentration of acid
${A^ - }$ is the concentration of conjugate base
This is a buffer solution that contains pyridine, $0.800\% {\text{ }}{{\text{C}}_5}{H_5}N$ , which is a weak base, and pyridinium chloride, $0.990\% {\text{ }}{{\text{C}}_5}{H_5}NHCl$, the salt of its conjugate acid, the pyridinium cation.
In order to use the Henderson - Hasselbalch equation, for a buffer that contains a weak base and its conjugate looks like this,
$pOH = p{K_b} + \log \left( {\dfrac{{conjugate{\text{ acid}}}}{{weak{\text{ }}base}}} \right)$
For this we need to determine the concentrations of pyridine and the pyridium cation. The value of the base dissociation constant which is denoted as ${K_b}$, for pyridine is

${K_b} = 1.7 \times {10^{ - 9}}$
Now, the density of the solution is not given in the problem, but however, since we are dealing with small amounts of pyridine and pyridinium chloride, we can assume that the density of the solution is equal to that of water.
Hence,
${\rho _{solution}} \approx {\rho _{water}} \approx 1g/mL$
Now, let us assume that we are dealing with a $1L$ sample of this buffer solution.
Since you know that
$1L = {10^3}mL$
we can say that this sample will be equivalent to
$1L \times \dfrac{{{{10}^3}mL}}{{1L}} = {10^3}ml$
Now this sample will have a mass of
${10^3}mL \times \dfrac{{1g}}{{1mL}} = {10^3}g$
The solution is said to be $0.800\% $ by mass pyridine and $0.990\% $ by mass pyridinium chloride.
We can use these concentrations to find the mass of the two chemical species in this sample
$\
  {10^3}g{\text{ solution}} \times \dfrac{{0.800g{\text{ }}{{\text{C}}_5}{H_5}N}}{{100g{\text{ }}solution}} = 8g{\text{ }}{{\text{C}}_5}{H_5}N \\
  {10^3}g{\text{ solution}} \times \dfrac{{0.990g{\text{ }}{{\text{C}}_5}{H_5}NHCl}}{{100g{\text{ }}solution}} = 9.90g{\text{ }}{{\text{C}}_5}{H_5}NHCl \\
\ $
Now we can use the molar masses of the two compounds to determine how many moles of each you have present
Molar mass for pyridine is $79.1g$
Molar mass of pyridinium chloride is $115.56g$
Hence,
$\
  8g \times \dfrac{{1mole{\text{ }}{{\text{C}}_5}{H_5}N}}{{79.1g}} = 0.10114moles{\text{ }}{{\text{C}}_5}{H_5}N \\
  9.90g \times \dfrac{{1mole{\text{ }}{{\text{C}}_5}{H_5}NHCl}}{{115.56g}} = 0.085670moles{\text{ }}{{\text{C}}_5}{H_5}NHCl \\
\ $
Pyridinium chloride dissociates in a $1:1$ mole ratio to form pyridinium cations and chloride anions
${C_5}{H_5}NHC{l_{(aq)}} \to {C_5}{H_5}N{H^ + }_{(aq)} + C{l^ - }_{(aq)}$
This means that for every mole of pyridinium chloride that is dissolved in the solution, we get one mole of pyridinium cations.
Since this sample has a total volume of $1L$, the molarity of the two species will be
\[\
  [{C_5}{H_5}N] = \dfrac{{0.10114{\text{ moles}}}}{{1L}} = 0.10114{\text{ }}M \\
  [{C_5}{H_5}N{H^ + }] = \dfrac{{0.085670{\text{ }}moles}}{{1L}} = 0.085670{\text{ }}M \\
\ \]
Now we use the Henderson - Hasselbalch equation to find the pOH of the buffer
\[pOH = - log({K_b}) + log\left( {\dfrac{{[{C_5}{H_5}N{H^ + }]}}{{{C_5}{H_5}N}}} \right)\]
Now by substituting the values we get,
\[\
  pOH = - log(1.7 \times {10^{ - 9}}) + log\left( {\dfrac{{0.085670M}}{{0.10114M}}} \right) \\
  pOH = 8.77 + ( - 0.0721) = 8.70 \\
\ \]
Since you know that at room temperature,
\[pH + pOH = 14\]
Hence the $pH$ of the solution is equal to
\[pH = 14 - 8.70 = 5.30\]

Note: We must note that if the concentrations of the weak acid and its conjugate base is high then the solution can be resistant to changes in hydrogen ion concentration or the $pH$. The change in $pH$ of a buffer solution when an acid or base is added can be calculated by combining the balanced equation for the reaction and the equilibrium acid dissociation constant.