Question

How would you use the Henderson–Hasselbalch equation to calculate the $pH$ of each solution?
Solution that contains $1.40$by mass and $1.18$by mass?

Answer 1

The Henderson-hasselbach equation the $pH$ of an equation is calculated if it contains buffer solution of the involved chemicals and if the dissociation constant of acid is represented as ${{K}_{a}}$, has a known value the $pH$ could be calculated with the formula.

Complete step-by-step answer: $pH$ is a measure of acidity or basicity of a solution which uses the concentration of protons in the solution to determine the acidity. The Henderson–Hasselbalch equation is an equation which relates the $pH$ of the solution along with the $p{{K}_{a}}$ , which is the ,measure of acidic strength of the solution, and we take the molar concentration as well.
In order to answer this question, we need to define the Henderson–Hasselbalch equation first which is,
$pH=p{{K}_{a}}+{{\log }_{10}}\left( \dfrac{[{{A}^{-}}]}{[HA]} \right)$
Where, $pH$ signifies the acidity of the buffer solution and $p{{K}_{a}}$ is the measure of acidity, which is also the negative logarithm of the dissociation constant. $[HA]$ is the concentration of the acid which we would be considering and $[{{A}^{-}}]$ is the concentration of the conjugate base which we would obtain if the acid loses its proton in the solution.
Now, we will consider the solution of ethylamine at first. Since it is a basic buffer, the equation would become,
$pOH=p{{K}_{b}}+log(\dfrac{[B{{H}^{+}}]}{[B]})$
Where, $pOH$ signifies the acidity of the buffer solution and $p{{K}_{b}}$ is the measure of acidity, which is also the negative logarithm of the dissociation constant. $[B{{H}^{+}}]$ is the concentration of the conjugate acid which we would be considering and $[B]$ is the concentration of base.
In the case of ethylamine, the value of $p{{K}_{b}}$ would be $3.19$.
Now, let us assume we take a thousand gram of solution, then it would contain $14.0g$ of ${{C}_{2}}{{H}_{5}}N{{H}_{2}}~$ which is the $[B]$, and $11.8g$ of ${{C}_{2}}{{H}_{5}}N{{H}_{3}}Br~$ which is the $[B{{H}^{+}}]$ in the solution. Now we would calculate the number of moles of both separately.
$Moles{ }of{ }B=14.0g{ }B\times \dfrac{1{ }mol{ }B}{45.08g{ }B}=0.3106{ }mol{ }B$
As we can see we used the definition of the number of moles for the calculation which is, number of moles is equal to the mass present per molar mass of that substance. And molar mass of ${{C}_{2}}{{H}_{5}}N{{H}_{2}}~$ is $45.08g$. Now, we will calculate the number of moles of ${{C}_{2}}{{H}_{5}}N{{H}_{3}}Br~$ in the same way.
$Moles{ }B{{H}^{+}}=11.8g{ }B{{H}^{+}}\times \dfrac{1{ }mol{ }B{{H}^{+}}}{126.00g{ }B{{H}^{+}}}=0.093{ }65{ }mol{ }B{{H}^{+}}$
So, the number of moles of ${{C}_{2}}{{H}_{5}}N{{H}_{3}}Br~$ came out to be, $0.093{ }65{ }mol$
Now, we would substitute these values in the equation, and we get,
$pOH=p{{K}_{b}}+log(\dfrac{[B{{H}^{+}}]}{[B]})=3.19+log(\dfrac{0.093{ }65mol}{0.3106mol})=3.19+log(0.3015)$
$=3.19{ }{ }0.52=2.67$
Since we know the value of $pOH$, we can easily find out the value of $pH$ by using the relation,
$pH=14.00{ }{ }pOH$
So, we get,
$=14.00{ }{ }2.67=11.33$
Which is the required answer.

Note: The $pH$ of a solution can be calculated if the $pOH$ of the solution is known, by the logic that the sum of both would be equal to $14$ which is the maximum value of $pH$ scale.
The Henderson–Hasselbalch equation, establishes a relationship between $pH$ or $pOH$ of the solution with the $p{{K}_{a}}$ or $p{{K}_{b}}$ of acid or base along with the concentration of corresponding acid or bases.