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Question: 1M weak monoacidic base (BOH) solution is diluted by 100 times. The pH change of the solution is: \(...

1M weak monoacidic base (BOH) solution is diluted by 100 times. The pH change of the solution is: (Kb=10−5)({{K}_{b}}={{10}^{-5}})
A. 2
B. -2
C. 1
D. -1

Explanation

Solution

A strong base is considered that liberates more OH ions, while a weak base liberates less OH ions. base dissociation constant, tells us the ability of a base to impart hydroxide ions in an aqueous solution. It is denoted as Kb{{K}_{b}}.

Formula used: pH for a weak base, pH = 12(pKb−log⁡C)\dfrac{1}{2}\left( p{{K}_{b}}-\log C \right) where C is concentration.

Complete step by step solution: We have been given a weak base with concentration 1 M, which is diluted 100 times. We have to calculate the pH change of the solution. Given the base dissociation constant, (Kb=10−5)({{K}_{b}}={{10}^{-5}})
Since, the base, BOH is of 1 M, means it contains 1 mole of B -OH particles in a volume of 1 L solution. When this solution is diluted 100 times, the final volume of the solution came out to be 100 L. So, 1 M of BOH in 100 L of solution will be concentration of OH as,
[OH−]=1100=10−2[O{{H}^{-}}]=\dfrac{1}{100}={{10}^{-2}}, so C = 10−2{{10}^{-2}}
Keeping this value of concentration in the formula for pH of a weak base, we have,
pH=12[pKb−log⁡10−2]pH=\dfrac{1}{2}\left[ p{{K}_{b}}-\log {{10}^{-2}} \right]
pH = pKb+22\dfrac{p{{K}_{b}}+2}{2}
pH = 1

Hence, change in pH of this 100 times diluted solution is calculated to be 1, thus option C is correct.

Note: The initial pH is taken to be pH=12[pKb]pH=\dfrac{1}{2}\left[ p{{K}_{b}} \right], therefore the change in pH came out to be 1. As, pKbp{{K}_{b}}is dissociation constant for a base, dissociation constant for an acid is pKap{{K}_{a}}. For a weak acid the same formula is used for the pH as, 12(pKa−log⁡C)\dfrac{1}{2}\left( p{{K}_{a}}-\log C \right)