Finding final pH of the buffer solution without applying Henderson equation

Question

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).

Answer 1

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:

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

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

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

Answer 2

There is an accepted answer for this question. I think there is an another way to solve the problem using basic principles.

Beaker A contains $\pu{25 cm^3}$ of $\pu{0.1 M}$ $\ce{NH3}$ solution while beaker B contains $\pu{5 cm^3}$ of $\pu{0.1 M}$ $\ce{HCl}$ solution. Since $\ce{HCl}$ is a strong acid, all of $\ce{HCl}$ ($\pu{0.1 \times 0.002 mol} = \pu{0.0002 mol}$) in the beaker B has been totally dissociated:

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

Thus, when the solutions in A and B are mixed, $\pu{0.0002 mol}$ of $\ce{NH3}$ will react with $\ce{HCl}$ to give $\pu{0.0002 mol}$ of $\ce{NH4Cl}$, which is also completely ionized in final solution:

$$\ce{NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)} \tag1$$

Since final volume of of combined solutions is $\pu{0.027 L}$, initial concentrations in final solution are: $$[\ce{NH4+}] = [\ce{Cl-}] = \frac{\pu{0.0002 mol}}{\pu{0.027 L}} = \pu{0.00741 M} $$ $$[\ce{NH3}] = \frac{\left(\pu{0.0025 mol} - \pu{0.0002 mol}\right)}{\pu{0.027 L}} = \pu{0.0852 M} $$

The amount of $\ce{NH3}$ remain in the final solution $(\pu{0.0023 mol})$ would be partially dissociated according to its $K_\mathrm{b} $:

$$\ce{NH3 + H2O <=> NH4+ + OH- } \quad \quad K_\mathrm{b} = \pu{3.3E-5}\tag2$$

If the amount of $[\ce{NH3}]$ dissociated to receive the equlibriun is $\alpha \ \pu {M}$, then the final concentrations at equilibrium are: $[\ce{NH3}] = \pu{(0.0852 - \alpha) M},$ $[\ce{NH4+}] = \pu{(0.00741 + \alpha) M},$ and $[\ce{OH-}] = \pu{(\alpha) M}.$ Apply these values to $K_\mathrm{b} = \pu{3.3E-5}$ for the equation $(2)$:

$$K_\mathrm{b} = \pu{3.3E-5} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{(0.00741 + \alpha)(\alpha)}{(0.0852 - \alpha)} \tag3$$

It is safe to assume $\alpha \lt \lt 0.0852,$ the equation $(3)$ can be rewritten as:

$$\alpha^2 + 0.00741 \alpha - 0.0852 \times \pu{3.3E-5} = 0\tag4$$ When solve for $\alpha$: $$\alpha = \pu{3.89E-4}$$

Since $\alpha = [\ce{OH-}]$, $\mathrm{pOH} = -\log (\pu{3.89E-4}) = 3.41$ Thus, the $\mathrm{pH}$ at equilibrium would be:

$$\mathrm{pH} + \mathrm{pH} = 14.0 \ \ \Rightarrow \ \ \mathrm{pH} = 14.0 - 3.41 = 10.59 $$

Note that the equation $(4)$ could be simplify by further assuming $\alpha \lt \lt 0.00741 = \pu{7.41E-3}.$ However, you can see that assumption is not entirely correct by looking at calculated $\alpha = \pu{3.89E-4}$.

For comparison, let's see what is the answer if you have used Henderson-Hasselbalch equation for the weak base $\ce{NH3}$:

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log \left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right) = -\log (\pu{3.3E-5}) + \log \left(\frac{\pu{0.00741 M}}{\pu{0.0852 M}}\right) = 3.42 $$

Thus, $\mathrm{pH} = 14.0 - 3.42 = 10.58 $.