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Question: The \[pH\] of a 0.1 \[M\] solution of \[N{H_4}OH\] (having \[{K_b} = 1.0 \times {10^{ - 5}}\] ) is e...

The pHpH of a 0.1 MM solution of NH4OHN{H_4}OH (having Kb=1.0×10−5{K_b} = 1.0 \times {10^{ - 5}} ) is equal to:
A) 1010
B) 66
C) 1111
D) 1212

Explanation

Solution

We must remember that the pHpH value of a solution may be defined as the negative logarithm of base 10 of the [H3O+][{H_3}{O^ + }] concentration which is expressed in moles per liter. The pOHpOH value of a solution may be defined as the negative logarithm of base 10 of the [OH−][O{H^ - }] present in the solution. The pHpH and the pOHpOH are related to each other as follows:pH+pOH=14pH + pOH = 14 .

Complete step by step answer:
Now we can use ammonium hydroxide to determine what concentration of hydroxide anions OH−O{H^ - } would cause the solid to precipitate out of solution. As we know the dissociation equilibrium for ammonium hydroxides will be given as follows
NH4OH→NH4++OH−N{H_4}OH \to N{H_4}^ + + O{H^ - }
NH4OHN{H_4}OH is a weak base. Here we have the value of Kb=1.0×10−5{K_b} = 1.0 \times {10^{ - 5}}
To find the concentration of the hydroxide anions would be as,
[OH−]=Cα=C×Kb[O{H^ - }] = C\alpha = \sqrt {C \times {K_b}}
Given data: Kb=1.0×10−5{K_b} = 1.0 \times {10^{ - 5}} and C=0.1MC = 0.1M
Substituting these values in formula we get,
[OH−]=0.1×(1×10−5)[O{H^ - }] = \sqrt {0.1 \times (1 \times {{10}^{ - 5}}} ) or
[OH−]=10−1×(1×10−5)\left[ {O{H^ - }} \right] = \sqrt {{{10}^{ - 1}} \times \left( {1 \times {{10}^{ - 5}}} \right)}
⇒\Rightarrow [OH−]=10−6\left[ {O{H^ - }} \right] = \sqrt {{{10}^{ - 6}}}
Cancelling the root square we get,
\Rightarrow $$$[O{H^ - }] = {10^{ - 3}}M$$ On using the concentration of hydroxide anions to find the solution, $$pOH = - \log [O{H^ - }]$$ \Rightarrow $pOH=−log⁡[10−3]pOH = - \log [{10^{ - 3}}]
As we know,
KW=[H+][OH−]{K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]
On substituting the values we get,
10−14=[H+][10−3]{10^{ - 14}} = \left[ {{H^ + }} \right]\left[ {{{10}^{ - 3}}} \right]
On simplification we get,
[H+]=10−11\left[ {{H^ + }} \right] = {10^{ - 11}}
As we know that,
pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
Now we can substitute the known value we get,
−log⁡(1×10−11)=11- \log \left( {1 \times {{10}^{ - 11}}} \right) = 11
Therefore, the option C is correct.

Note:
We must know that the pHpH of a neutral solution is 7. The pHpH value decreases as the solution becomes increasingly acidic. The solutions with the pHpH range 0-2 are strongly acidic those with pHpH 2-4 are moderately acidic while the solutions with pHpH values between 4-7 are weakly acidic. As the pHpH value increases the solution gradually becomes more basic. The solutions having a pHpH range of -10 are weakly basic; those having the range between 10-12 are moderately basic whereas the solution having pHpH a range between 13-14 is highly basic.