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# How would you calculate the pH of the 0.39 M $N{{H}_{3}}$/ 0.73 M $N{{H}_{4}}Cl$buffer system?

Last updated date: 12th Aug 2024
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Hint:A buffer solution or the pH buffer or we can say the hydrogen ion buffer is considered as the aqueous solution. It consists of the mixture of the conjugate base and the weak acid or vice versa. The change in its pH is very little when the addition of a small amount of either the strong base or strong acid is added to it. Buffer solutions are used in keeping the pH at a nearest constant value in the wide range of chemical applications.

Complete step-by-step answer: So in this question we willl use the Henderson Hasselbalch equation. The equation is the following for the base and its conjugate acid:
$pOH=p{{K}_{b}}+\log \frac{[B{{H}^{+}}]}{[B]}$ $pH=14-5.01=8.99$
Here we represent $B{{H}^{+}}$as the ammonium which act as the Bronsted acid or the proton donor for the equilibrium process which is described as the reaction below:
$N{{H}_{4}}^{+}(aq)+{{H}_{2}}O(l)N{{H}_{3}}(aq)+O{{H}^{-}}(aq)$
We know that the ${{K}_{b}}$ for ammonia is equal to $1.8\times {{10}^{-5}}$. Now we will calculate the $p{{K}_{b}}$for ammonia by the following formula:
$p{{K}_{b}}=-\log ({{K}_{b}})$
Now we will calculate the $p{{K}_{b}}$by substituting the values in the above formula:
$p{{K}_{b}}=-\log (1.8\times {{10}^{-5}})=4.74$
Now we will calculate the pOH by the above formula by substituting the values in it. We know$B{{H}^{+}}$ is equal to 0.73 and B is equal to 0.39. now let us substitute in the formula:
$pOH=4.74+\log \frac{[0.73]}{[0.39]}$
On simplifying further we get:
$pOH=5.01$
Now we will calculate the pH by the following formula:
$pH=14-pOH$
Substituting the value of pOH we get,
$pH=14-5.01=8.99$
So the pH of this buffer system is 8.99.

Note:pH is the potential of hydrogen or the power of hydrogen. The pH ranges from 0 to 14. Below 7 the solution is acidic, at seven the solution is neutral and above 7 it is basic. It helps in measuring the alkalinity and acidity of the solution. The base and the conjugate acid of it or the acid and the conjugate base of it react or participate in the equilibrium constant with each other.