# Finding final pH of the buffer solution without applying Henderson equation

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).

• 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?
– MaxW
Nov 2, 2020 at 11:02
• 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? Nov 2, 2020 at 11:19
• No way the pH could be acidic given the problem statement so a pH of 1.819 is ridiculous.
– MaxW
Nov 2, 2020 at 11:27
• 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.
– MaxW
Nov 2, 2020 at 11:37
• In my understanding, the task author wanted to force you to solve the task by alternative way, not by the H. equation. Nov 2, 2020 at 15:54

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:

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

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

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

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