Question

The \[pH\] of a 0.1 \[M\] solution of \[N{H_4}OH\] (having \[{K_b} = 1.0 \times {10^{ - 5}}\] ) is equal to:
A) $10$
B) $6$
C) $11$
D) $12$

Answer 1

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

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

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