Question

Consider several \(25.00-\mathrm{mL}\) solutions of perchloric acid. What is the molarity of the acid solution neutralized by (a) \(17.25 \mathrm{~mL}\) of \(0.3471 \mathrm{M}\) ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\). (b) \(14.17 \mathrm{~g}\) of strontium hydroxide. (c) \(41.73 \mathrm{~mL}\) of an \(18 \%\) (by mass) solution of ammonia \((d=0.9295 \mathrm{~g} / \mathrm{mL})\)

Short answer

Question: Determine the molarity of a perchloric acid (HClO4) solution in three different scenarios where it is neutralized by (a) ethylamine, (b) strontium hydroxide, and (c) ammonia solution. Answer: The molarity of the perchloric acid solution is (a) 0.2393 M when neutralized by ethylamine, (b) 9.312 M when neutralized by strontium hydroxide, and (c) 16.48 M when neutralized by ammonia solution.

Step-by-step solution

(a) Balancing the reaction equation

First, let's write the balanced chemical reaction equation for the neutralization of perchloric acid (HClO4) and ethylamine (C2H5NH2): $$ \mathrm{HClO}_{4}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{4} $$ The balanced equation shows that there is a 1:1 relationship between the moles of perchloric acid and ethylamine.

(a) Moles of ethylamine

Now, let's calculate the moles of ethylamine using the given volume and molarity: $$ \text{moles of ethylamine} = 0.01725 \,\text{L} \times 0.3471\, \text{M} = 0.005983\, \text{mol} $$

(a) Moles of perchloric acid

Since the stoichiometry of the reaction between perchloric acid and ethylamine is 1:1, the moles of perchloric acid are equal to the moles of ethylamine: $$ \text{moles of perchloric acid} = 0.005983\, \text{mol} $$

(a) Molarity of perchloric acid

Finally, we find the molarity of perchloric acid using the volume and moles of acid: $$ \mathrm{M} =\frac{\text{moles}}{\text{volume in liters}} = \frac{0.005983\,\text{mol}}{0.025\,\text{L}} = 0.2393\,\mathrm{M} $$ Therefore, the molarity of the perchloric acid solution neutralized by the ethylamine is \(0.2393\,\mathrm{M}\).

(b) Balancing the reaction equation

For the neutralization of perchloric acid (HClO4) and strontium hydroxide (Sr(OH)2), the balanced chemical reaction equation is: $$ 2\,\mathrm{HClO}_{4} + \mathrm{Sr(OH)_{2}} \rightarrow \mathrm{Sr(ClO_{4})_{2}} + 2\,\mathrm{H_{2}O} $$

(b) Moles of strontium hydroxide

First, we need to find the moles of strontium hydroxide given its mass (14.17g) and molar mass (121.63 g/mol): $$ \text{moles of Sr(OH)2} = \frac{14.17\, \text{g}}{121.63 \,\text{g/mol}} = 0.1164\, \text{mol} $$

(b) Moles of perchloric acid

Since the stoichiometry of the reaction between perchloric acid and strontium hydroxide is 2:1, we can find the moles of perchloric acid: $$ \text{moles of perchloric acid} = 2 \times \text{moles of Sr(OH)2} = 2 \times 0.1164\, \text{mol} = 0.2328\, \text{mol} $$

(b) Molarity of perchloric acid

We can now calculate the molarity of the perchloric acid solution using the volume and moles of acid: $$ \mathrm{M} =\frac{\text{moles}}{\text{volume in liters}} = \frac{0.2328\,\text{mol}}{0.025\,\text{L}} = 9.312\,\mathrm{M} $$ Thus, the molarity of the perchloric acid solution neutralized by strontium hydroxide is \(9.312\,\mathrm{M}\).

(c) Balancing the reaction equation

For the neutralization of perchloric acid (HClO4) and ammonia (NH3), the balanced chemical reaction equation is: $$ \mathrm{HClO}_{4}+\mathrm{NH}_{3} \rightarrow \mathrm{NH}_{4}\mathrm{ClO}_{4} $$

(c) Mass and moles of ammonia

First, let's find the mass and moles of ammonia in the given solution. The density of the solution is given as 0.9295 g/mL, and the mass percentage of ammonia is 18%. We calculate the mass of ammonia in grams: $$ \text{mass of ammonia} = \text{volume} \times \text{density} \times \frac{\text{mass percentage}}{100} = 41.73\,\text{mL} \times 0.9295\, \frac{\text{g}}{\text{mL}} \times \frac{18}{100} = 7.012\,\text{g} $$ Next, we find the moles of ammonia using its molar mass (17.03 g/mol): $$ \text{moles of ammonia} = \frac{7.012\, \text{g}}{17.03 \,\text{g/mol}} = 0.4119\, \text{mol} $$

(c) Moles of perchloric acid

Since the stoichiometry of the reaction between perchloric acid and ammonia is 1:1, the moles of perchloric acid are equal to the moles of ammonia: $$ \text{moles of perchloric acid} = 0.4119\, \text{mol} $$

(c) Molarity of perchloric acid

Lastly, we determine the molarity of the perchloric acid solution using the volume and moles of acid: $$ \mathrm{M} =\frac{\text{moles}}{\text{volume in liters}} = \frac{0.4119\,\text{mol}}{0.025\,\text{L}} = 16.48\,\mathrm{M} $$ So, the molarity of the perchloric acid solution neutralized by the ammonia solution is \(16.48\,\mathrm{M}\).