Question

The pH of a $ 0.1M $ $ N{H_3} $ ​ solution ( $ {K_b} = 1.8 \times {10^{ - 5}} $ ) is:
(A) $ 11.3 $
(B) $ 1 $
(C) $ 13 $
(D) None of the above

Answer 1

Hint: The molarity is used as the concentration here. The pH of a solution is expressed as the negative logarithm of the concentration of hydrogen ions. We shall calculate the concentration of hydroxide ions using the formula given and use it to find the pOH and thus, the pH.

Formula Used: $ \left[ {O{H^ - }} \right] = \sqrt {{K_b} \times c} $
 $ pH = - {\text{log}}\left[ {{H^ + }} \right] $

Complete step by step solution:
It has been given that the concentration of ammonia is $ 0.1M $ and $ {K_b} = 1.8 \times {10^{ - 5}} $ where $ {K_b} $ is the equilibrium constant for the ionization of a base.
pH is the quantitative measure of the acidity or basicity of aqueous or other liquid solutions. translates the values of the concentration of the hydrogen ion—which ordinarily ranges between about 1 and $ {10^{ - 14}} $ gram-equivalents per litre—into numbers between 0 and 14. In pure water, which is neutral (neither acidic nor alkaline), the concentration of the hydrogen ion is $ {10^{ - 7}} $ gram-equivalents per litre, which corresponds to a pH of 7. A solution with a pH less than 7 is considered acidic; a solution with a pH greater than 7 is considered basic, or alkaline.
We know, to calculate the $ pH $ of an aqueous solution we need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression:
 $ pH = - {\text{log}}\left[ {{H^ + }} \right] $
We know that, $ \left[ {O{H^ - }} \right] = \sqrt {{K_b} \times c} $ where $ c $ is the concentration.
Assigning the values to the respective variables, we can write,
 $ = \sqrt {1.8 \times {{10}^{ - 5}} \times 0.1} = 1.34 \times {10^{ - 3}} $ .
 From $ \left[ {O{H^ - }} \right] $ , the $ p\left[ {O{H^ - }} \right] $ value can be calculated by, $ p\left[ {O{H^ - }} \right] = - \log \left[ {O{H^ - }} \right] $ .
Thus, $ p\left[ {O{H^ - }} \right] = - \log \left( {1.34 \times {{10}^{ - 3}}} \right) $ .
Simplifying this equation by using the laws of logarithm, we get,
 $ p\left[ {O{H^ - }} \right] = 3 - \log \left( {1.34} \right) $
The value of $ \log \left( {1.34} \right) \simeq 1.271 $ is put in the equation.
  $ p\left[ {O{H^ - }} \right] = 3 - 1.271 = 2.87 $ .
Thus, using this value in the equation for autoionization of water, we get,
 $ p\left[ {{H^ + }} \right] + p\left[ {O{H^ - }} \right] = 14 $
Using the value, $ p\left[ {O{H^ - }} \right] = 2.87 $ , we get,
 $ p\left[ {{H^ + }} \right] + 2.87 = 14 $ .
 $ p\left[ {{H^ + }} \right] = 11.13 $
Hence, option A is the correct answer.

Note:
An equilibrium is established between the ions produced and the unionized water. An equilibrium expression can be written for this system.
 $ {K_w} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right] $
 $ {K_w} $ is used to represent the equilibrium constant for the ionization of water. The equilibrium constant for the autoionization of water, $ {K_w} $ is $ {10^{ - 14}} $ $ {25^ \circ }C $ . Thus, $ p{K_w} = - \log \left( {{{10}^{ - 14}}} \right) $ . Simplifying, $ p{K_w} = 14 $ .