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Question: The volume of one drop of aqueous solution from an eyedropper is approximately 0.05 mL. One such dro...

The volume of one drop of aqueous solution from an eyedropper is approximately 0.05 mL. One such drop of 0.2 M HCl is added to 100 mL of distilled water. The pH of the resulting solution will be
(A)- 4.0
(B)- 7.0
(C)- 3.0
(D)- 5.5

Explanation

Solution

The given solution is obtained by the dilution of the aq. HCl solution by adding to a large volume of water, without the number of moles of HCl remaining unchanged in the resulting solution.

Complete step by step answer:
Given the volume, V1{{V}_{1}} and concentration, M1{{M}_{1}} of one drop of aqueous hydrochloric solution to be 0.05 mL and 0.2 M respectively.
When aqueous solution from the dropper is mixed into the distilled water of volume, V2{{V}_{2}} . That is the dilution of the aqueous HCl solution as it is added to a large volume of solvent. But because the moles of HCl solute remain equal in both the concentrated and diluted solution, we obtain the equation of dilution. Then, the concentration of the solution,M2{{M}_{2}}will be obtained by the dilution equation relation, as follows:
M1V1=M2V2{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}
Substituting the values given value, we will obtain concentration of the diluted solution, M2{{M}_{2}} as:
0.2×0.05=M2×1000.2\times 0.05={{M}_{2}}\times 100
M2=0.2×0.05100=1×10−4M{{M}_{2}}=\dfrac{0.2\times 0.05}{100}=1\times {{10}^{-4}}M
So, the concentration of the resulting solution is 1×10−4M1\times {{10}^{-4}}M. There is a decrease in the concentration of the HCl as the number of ions present per unit volume has decreased on dilution. Also, HCl being a strong acid, it dissociates completely on dissolving in water to produce H+ and Cl−{{H}^{+}}\,and\,C{{l}^{-}} ions in it. Then, we can find the pH of the resulting solution with respect to the H+{{H}^{+}} ions. Since, the concentration of the solution M2{{M}_{2}} is equal to the concentration of [H+]\left[ {{H}^{+}} \right].
So, we have the relation as follows:
pH=−log⁡[H+]pH=-\log \left[ {{H}^{+}} \right]
pH=−log⁡[10−4]pH=-\log \left[ {{10}^{-4}} \right]
pH=−(− 4)log⁡(10)=4pH=-(-\,4)\log (10)=4

The pH of the resulting solution will be option (A)- 4.0.

Note: Through the process of dilution we obtain the solution of the desired concentration. Also, in the dilution equation, three quantities are known. Only then, the unknown quantity can be calculated.