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A is beaker with $\pu{0.1 M}$ $\pu{25 cm^3}$ solution of $\ce{NH3}.$ B is beaker with $\pu{0.1 M}$ $\pu{5 cm^3}$ $\ce{HCl}.$ Now the solution of A and B is mixed. What will be the final $\mathrm{pH}$ of the mixture? $K_\mathrm{b} = \pu{3.3E-5}.$

I believe the mixture will become a buffer solution and we should use Henderson equation to solve this.

$$n(\ce{NH3}) = \pu{0.1 mol L^-1} × (\pu{25E-3 L}) = \pu{2.5E-3 mol}\tag{1}$$

$$n(\ce{HCl}) = \pu{0.1 mol L^-1} × (\pu{5E-3 L}) = \pu{5E-4 mol}\tag{2}$$

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log\frac{n(\ce{NH4Cl})}{n(\ce{NH4OH})}\tag{3}$$

$$\ce{NH3 + HCl -> NH4Cl}$$

$$n(\ce{HCl}) = n(\ce{NH4Cl}) = \pu{5E-4 mol}\tag{4}$$

$$n(\ce{NH3}) = n(\ce{NH4OH}) = \pu{2.5E-3 mol} - \pu{5E-4 mol} = \pu{2E-3 mol}\tag{5}$$

$$ \begin{align} \mathrm{pOH} &= \mathrm{p}K_\mathrm{b} + \log\frac{\pu{5E-4 mol}}{\pu{2E-3 mol}}\\ &= 4.4819 - 0.602\\ &\approx 3.88 \tag{6} \end{align}$$

$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 3.88 = 10.12\tag{7}$$

However, the book from which I am solving this problem suggests that this problem should not be solved by Henderson equation without providing any reason. Can anyone tell me why this solution should not be considered a basic buffer? I would really appreciate some opinion on this.

P.S. The problem was given in Bangla. I tried my best to translate it. This is an admission question of KUET-2019 Bangladesh. My textbook has this question, but I'm not sure about the solution. Please note that this is an admission question of previous year (2019).

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    $\begingroup$ The HH equation should yield a reasonable result in this case. The precision of the measurements is sloppy. // One reason to solve without HH equation would be to see if you understood how to solve the general problem without the HH assumption. Can you solve problem without using HH? $\endgroup$ – MaxW Nov 2 '20 at 11:02
  • $\begingroup$ No, I can't solve the problem without HH assumption. The solution in my textbook says that the pH should be 1.819 which is way of what I got. Can you please explain how this could be solved without HH assumption? $\endgroup$ – Sadatul islam Sadi Nov 2 '20 at 11:19
  • $\begingroup$ No way the pH could be acidic given the problem statement so a pH of 1.819 is ridiculous. $\endgroup$ – MaxW Nov 2 '20 at 11:27
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    $\begingroup$ You can't have an appreciable amount of $\ce{H+}$ floating around in an ammonia solution. So essentially all (to two significant figures) the $\ce{HCl}$ reacts to make $\ce{NH4+}$. However some of the $\ce{NH3}$ will react with water to make $\ce{NH4+}$ too. So the solution should be slightly more basic that what you'd get from HH equation. $\endgroup$ – MaxW Nov 2 '20 at 11:37
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    $\begingroup$ In my understanding, the task author wanted to force you to solve the task by alternative way, not by the H. equation. $\endgroup$ – Poutnik Nov 2 '20 at 15:54
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As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$

$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$

from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:

$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$

$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$

$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

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