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

Hint: 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 ${{K}_{b}}$.

Formula used: pH for a weak base, pH = $\dfrac{1}{2}\left( p{{K}_{b}}-\log C \right)$ 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, $({{K}_{b}}={{10}^{-5}})$
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,
\[[O{{H}^{-}}]=\dfrac{1}{100}={{10}^{-2}}\], so C = ${{10}^{-2}}$
Keeping this value of concentration in the formula for pH of a weak base, we have,
\[pH=\dfrac{1}{2}\left[ p{{K}_{b}}-\log {{10}^{-2}} \right]\]
pH = $\dfrac{p{{K}_{b}}+2}{2}$
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 \[pH=\dfrac{1}{2}\left[ p{{K}_{b}} \right]\], therefore the change in pH came out to be 1. As, $p{{K}_{b}}$is dissociation constant for a base, dissociation constant for an acid is $p{{K}_{a}}$. For a weak acid the same formula is used for the pH as, $\dfrac{1}{2}\left( p{{K}_{a}}-\log C \right)$