If the pH of aqueous solutions A, B, C and D are 1.9, 2.5, 2.1 and 3.0 then what will be the order of acidity? Select the proper choice from the given multiple choices.
(A) A
(B) D
(C) D
(D) D>C>B>A
Answers
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (Kw). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm:
pX = −log X
The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:
pH = −log[H3O+]
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
[H3O+] = 10−pH
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
pOH = −log[OH−]
or
[OH−] = 10−pOH
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the Kw expression:
Kw = [H3O+][OH−]
−logKw = −log([H3O+][OH−]) = −log[H3O+] + −log[OH−]
pKw = pH + pOH
At 25 °C, the value of Kw is 1.0 × 10−14, and so:
14.00 = pH +pOH
As we learned earlier, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 × 10−7 M at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
pH = −log[H3O+] = −log(1.0 × 10−7) = 7.00
pOH = −log[OH−] = −log(1.0 × 10−7) = 7.00