Test: Ionisation of Water - JEE MCQ

We know that pH equals to - log[H⁺] or the value of pH is inversely proportional to the concentration of H⁺.  So more the value of pH, less the solution be acidic.

First off, since NaOH is a strong base, it will dissociate completely into Na+ and OH-. Thus, we know that we have 0.01 M OH-.

However, we do not know anything about the concentration of H+. Fortunately, we do not need to, as pH + pOH = 14. So, if we find pOH, we can solve for pH. p is a mathematical function equivalent to -log. So, pH actually means -log[H+] (Note that brackets indicate concentration of).

pOH = -log 0.01M OH-

pOH = 2

pH + 2 = 14

pH = 12

This result makes sense, since a solution of strong base should have a high pH.

Since the acid is only 0.059% ionized, therefore the concentration of ions in solution = 0.1 x 0.059 / 100 = 0.000059

Ka = [H⁺] [X^–] / [HX] = (0.000059)² / 0.1 = 3.5 x 10^-8

The correct answer is Option D.
The concentration of OH- will always equal the concentration of H3O+.
PV = nRT, where R is a constant, and n is the quantity of gas measured in mols. The pH of a substance is related to the concentration of H+,
so if we raise or lower the Temperature or Pressure of pure water, we will alter the Volume (or concentration) of H3O+ and OH- ions. As we alter the concentrations we alter the pH and pOH. The only answer that is true no matter what the volume or concentration may be is D
 

If we use the relation, pH = – log [H3O⁺], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when H⁺ concentrations from acid and water are comparable, the concentration of H⁺ from water cannot be neglected.

Therefore,