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The solution with the lowest pH value is…

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The solution with the lowest pH value is

A. 5 ml of ?/10  HCL

B. 10 ml of ?/10  HCL

C. 15 ml of ?/5  HCL

D. 20 ml of ?/8  HCL




Be careful, I have seen brilliant people fail this…

If the capital letter M stands for molarity or molar concentration as is usually the case in most chemistry texts, then option C should be the answer. I will explain shortly…

The PH value of an acid is calculated with the formula:

PH = -log₁₀[H]

Where [H] is the molarity M, or molar concentration of hydrogen ion in the solution.

At this point, please note that the molar concentration of an acid will not change even if you reduce or halve the solution. For instance, 20 liters of an acid whose molarity is 5M will still maintain this molarity even if the volume of the solution is reduced, this is because molarity is all about the number of moles of the acid particles in one liter of the solution. So, halving or sharing the solution will not change its concentration and PH, each portion of the solution will still maintain the same molarity. The concentration will only change if water or another substance is added to it.

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So, from the question options, the volumes given in ml are inconsequential, only the molarity values M, are important.

So, solving for the PH of each to determine the lowest value,

The solution with the lowest pH value is

Now for the right answer to the above question:

  1. Option A is incorrect.
  2. Option B is incorrect. option B will basically have the same PH value as option A, the change in volume or quantity does not matter as long as they have the same molar concentration.
  3. C is correct.
  4. D is not the correct answer.


You may please note these/this:

  • Decreasing the volume of an acid does not change its molar concentration and by extension its PH value.
  • You do not reduce the concentration of an acid simply by halving or reducing its quantity.
  • Adding water or other substances is what can change its concentration.

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/ culled from 2015 JAMB-UTME Chemistry past question 22 /

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