# How Do I Determine The Strength Of Two Ammonia Solutions Using Titration Methods? [closed]

Two housewives disagreed on the strength of an ammonium cleaning agent. Housewife A thought that her brand was more effective than the brand used by housewife B. Plan and design an experiment that could be used to determine the more effective brand of cleaner.

I know that the aim is to determine which if the ammonium solutions are stronger, more concentrated (as the stronger ammonium solution would therefore be the more effective brand). So, a back titration must be used. $$\ce{HCl}$$ (hydrochloric acid) could be used in the titration.

However, I am not sure how the experiment would be carried out.

Assume concentration of solution A and B are $$a\%$$ and $$b\%$$, respectively (both are $$\%(w/v)$$). Suppose $$a\% \approx b\% \approx 5\%$$. Molar mass of ammonia ($$\ce{NH3}$$) is $$\pu{17.03 gmol-1}$$. Thus, if molarity of A and B are $$M_A$$ and $$M_B$$, respectively, then:

$$M_A = \frac{a \ \pu{g} \ \ce{NH3}}{\pu{100 mL}} \times \frac{\pu{1 mol}}{\pu{17.03 g} \ \ce{NH3}} \times \frac{\pu{1000 mL}}{\pu{1.0 L}} = \frac{10a}{17.03}\pu{molL-1}$$

Similarly, $$M_B = \dfrac{10b}{17.03}\pu{molL-1}$$.

Let's take $$\pu{100 mL}$$ of each solution in separate containers. Add to each $$\pu{100 mL}$$ of $$\pu{1.0 M}$$ $$\ce{HCl}$$ (assume volumes are additive). The following reaction would happen in each container:

$$\begin{array}{lccc} \ce{& NH3 &+ & HCl & -> & NH4Cl} \\ \text{I} & \frac{10a}{17.03}\times 0.10 && 1.0 \times 0.10 && 0 \\ \text{C} & -1.0 \times 0.10 && -1.0 \times 0.10 && +1.0 \times 0.10 \\ \text{E} & \frac{10a}{17.03}\times 0.10 - 1.0 \times 0.10 && 0 && +1.0 \times 0.10 \\ \end{array}$$

Therefore, at equilibrium in solution A, $$[\ce{NH3}] = \frac{\frac{10a}{17.03}\times 0.10 - 1.0 \times 0.10}{0.20} \ \pu{M}=\frac{10a-17.03}{34.06} \ \pu{M}$$.

And, $$[\ce{NH4Cl}] = \frac{1.0 \times 0.10}{0.20} \ \pu{M}= \pu{0.50 M}$$.

Similarly, at equilibrium in solution B, $$[\ce{NH3}] = \frac{\frac{10b}{17.03}\times 0.10 - 1.0 \times 0.10}{0.20} \ \pu{M}=\frac{10b-17.03}{34.06} \ \pu{M}$$.

And, $$[\ce{NH4Cl}] = \frac{1.0 \times 0.10}{0.20} \ \pu{M}= \pu{0.50 M}$$.

As a result, Solution A and B become buffer solutions after addition of $$\ce{HCl}$$. Now, we can measure $$\mathrm{pH}$$ of each solution. If both are in same strength, both buffer solutions give same $$\mathrm{pH}$$ value.

Suppose they have given different values. Suppose the values given are: $$\mathrm{pH}_{(A)} = 9.55$$ and $$\mathrm{pH}_{(B)} = 9.45$$. Assume for $$\ce{NH4+/NH3}$$ system, $$\mathrm{p}K_\mathrm{a} = 9.25$$

Now we can use the Henderson–Hasselbalch equation, $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\text{[base]}}{\text{[acid]}}\right)$$, to calculate each concentration (acid and base here are $$\ce{NH4Cl}$$ and $$\ce{NH3}$$, respectively):

$$9.55 = 9.25 + \log \left(\frac{\frac{10a-17.03}{34.06}}{0.50}\right) = 9.25 + \log \left(\frac{10a-17.03}{17.03}\right)\\ = 9.25 - 1.231 + \log (10a-17.03)$$

$$\log (10a-17.03) = 9.55 - 9.25 + 1.231 = 1.531 \\ \Rightarrow \ \therefore \ a = \frac{1}{10}(33.96 + 17.03) = 5.099$$

Similarly, for Buffer B:

$$9.45 = 9.25 + \log \left(\frac{\frac{10b-17.03}{34.06}}{0.50}\right) = 9.25 + \log \left(\frac{10b-17.03}{17.03}\right)\\ = 9.25 - 1.231 + \log (10b-17.03)$$

$$\log (10b-17.03) = 9.45 - 9.25 + 1.231 = 1.431 \\ \Rightarrow \ \therefore \ b = \frac{1}{10}(26.98 + 17.03) = 4.401$$

Thus, strength of the solution A is $$5.1\%$$ while that of the solution B is $$4.4\%$$. Clearly, solution A is stronger.

For housewives, a solution is simply to add Epsom salt in equal doses to each mix and let the $$\ce{Mg(OH)2}$$ settle out of solution.

Decant and repeat the experiment until one solution no longer produces any results of a white precipitate. The latter brand is weaker in its aqueous ammonia concentration based on the reaction:

$$\ce{\overset{Epsom salt}{MgSO4 · 7 H2O} + 2 NH3(aq) -> Mg(OH)2(s) + (NH4)2SO4(aq)}$$

To speed up the comparative test, one could experiment with a sample of each brand of ammonia which has been equally diluted with distilled water (record the respective volumes used for future tests).

For calculation purposes, I would start with an assumption that each solution is approximately $$5\%$$ aqueous ammonia.

• What was the point of switching from \underset to \overset in your edit? All style guides I know recommend to utilize space below for naming since the above area is reserved for charges, oxidation numbers and arrow pushing. If you edit the question solely for bumping it, note that this might be considered as abusive activity. – andselisk Jun 14 at 10:17
• I agree with @andselisk. Writing names above compounds looks so wrong and off-balance. – Nilay Ghosh Jun 14 at 11:05