Question

How much volume of 0.3M $N{{H}_{4}}OH$ should be mixed with 30mL 0.2M $N{{H}_{4}}Cl$ to make buffer of pH=12? (${{p}^{{{K}_{b}}}}$ $N{{H}_{4}}OH$=4.75)

Answer 1

Buffer of solution is made by mixing weak acid and its conjugate base or weak base and its conjugate acid (sometimes even salts). The answer here includes calculation of $pOH$ given by the formula, $pOH=p{{K}_{b}}+\log \dfrac{[salt]}{[base]}$

Complete step-by-step answer:
Let us take that V mL of $N{{H}_{4}}OH$ will be mixed with 30ml of $N{{H}_{4}}Cl$ to get buffer of pH=12
Thus, we get the total volume as V+30 mL
Now, from the previous lessons in chemistry we know that molarity of a compound is calculated to know the total number of moles of solute present per litre of solution .
Here it is as follows,
m mole for $N{{H}_{4}}OH$= 0.3$\times$V= 0.3V
Similarly, m mole for $N{{H}_{4}}Cl$ = 0.2$\times$30 $$$$
Now we must calculate the concentration of base and salt as per the formula.
Therefore, $[N{{H}_{4}}OH]=\dfrac{0.3V}{Totalvolume}$= $\dfrac{0.3V}{V+30}$
Also, $[N{{H}_{4}}Cl]=\dfrac{0.2\times 30}{V+30}$
To get pH=12, we need to mix base ($N{{H}_{4}}OH$) with salt($N{{H}_{4}}Cl$)
Thus pH of base =
$\Rightarrow$14-12= 4.75 + $\log \dfrac{(0.2\times 30)/(V+30)}{(0.3V)/(V+30)}$
[Since, pOH = 14-pH]
Therefore, 2-4.75= $\log \dfrac{(0.2\times 30)/(V+30)}{(0.3V)/(V+30)}$
-2.75= $\log \dfrac{6}{0.3V}$
$\Rightarrow$V= 112.5 mL
Thus, 112.5 mL of 0.3M $N{{H}_{4}}OH$ is to be mixed with 30 ml of 0.2M $N{{H}_{4}}Cl$ to get buffer of ph=12

Additional information: Acidity and basicity is measured on the basis of calculation of pH. Buffer solutions are used as a means of keeping pH at nearly a constant value in many chemical applications and experiments. These buffer solutions neutralize any added acid or base to maintain moderate pH making them a weaker acid or base. Interesting fact is that, body uses naturally synthesized buffer solutions inside it to maintain a constant pH.

Note: Do not get confused about which chemical is salt and which are acids or base. A basic definition of these would make the points clear. Another important point is that molarity and molality are two different quantities. Chances of getting confused about these may also lead to wrong answers.