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Question: What is the pH of a millimolar solution of ammonium hydroxide which is 20 % dissociated? (A) 3.699...

What is the pH of a millimolar solution of ammonium hydroxide which is 20 % dissociated?
(A) 3.699
(B) 10.301
(C) 4.691
(D) 9.301

Explanation

Solution

. We know that ammonium hydroxide is the weak base, therefore OH−O{H^ - } ions will dissociate partially. From the data given in the question we can calculate the concentration and degree of dissociation of OH−O{H^ - } ions. Later by using the relationship to pH and pOH we can easily calculate pH of a millimolar solution of ammonium hydroxide which is 20% dissociated.

Complete step by step answer:
In the question they have asked us to find the pH of millimolar solution of ammonium hydroxide which is 20 % dissociated.
Let us start finding the concentration of the NH4OHN{H_4}OH solution. It has been shown that NH4OHN{H_4}OH is present in millimolar concentration.
NH4OH=0.001MN{H_4}OH = 0.001M……. (1)
The ammonium hydroxide is a weak base, so NH4OHN{H_4}OH will dissociate as OH−O{H^ - } ions partially.
The degree of dissociation of NH4OHN{H_4}OH is 20 %
degree of dissociation =20%=20100=0.2 = 20\% = \dfrac{{20}}{{100}} = 0.2……. (2)
From the above value concentration of OH−O{H^ - }ions can be calculated
Concentration of OH -  = Concentration of solution×degree of dissociation{\text{Concentration of O}}{{\text{H}}^{\text{ - }}}{\text{ = Concentration of solution}} \times {\text{degree of dissociation}}
[OH−]=0.001×0.2=2×10−4[O{H^ - }] = 0.001 \times 0.2 = 2 \times {10^{ - 4}}…… (3)
We know that water has pOH = 7
Therefore, the concentration of OH−O{H^ - } ions will be
pOH=−log⁡[OH−]pOH = - \log [O{H^ - }]……... (4)
7=−log⁡[OH−]7 = - \log [O{H^ - }]
[OH−]=10−7[O{H^ - }] = {10^{ - 7}}….…... (5)

- To find the total concentration of OH−O{H^ - } we should add equation (3) and (5)
total[OH−]=(2×10−4)+10−7total [O{H^ - }] = (2 \times {10^{ - 4}}) + {10^{ - 7}}
total[OH−]=10−4(2+0.001)total [O{H^ - }] = {10^{ - 4}}(2 + 0.001)
total[OH−]=2.001×10−4total [O{H^ - }] = 2.001 \times {10^{ - 4}}
- Substituting [OH−][O{H^ - }] in equation (4)
pOH=−log⁡(2.001×10−4)pOH = - \log (2.001 \times {10^{ - 4}})
pOH=3.6988pOH = 3.6988
pH can be calculated using the equation
pH=14−pOHpH = 14 - pOH…… (6)
pH=14−3.6988pH = 14 - 3.6988
pH=10.301pH = 10.301
The pH of a millimolar solution of ammonium hydroxide which is 20 % dissociated is 10.301.
So, the correct answer is “Option B”.

Additional information:
Acids and bases are measured using pH or pOH scale. This scale provides the measure of H+{H^ + } or OH−O{H^ - } ion concentration.
- When we measure acids, then pH scale will give values less than 7, while pOH scale will give values greater than 7.
- When we measure bases, then pH scale will give values greater than 7, while pOH scale will give values less than 7.

Note: After getting the value of pOH we should not conclude the answer. Since in question they have asked for the pH value. pH and pOH are not the same. So, we should not get confused with that. We should be using that pOH value and substitute it in the equation (6) in order to obtain the pH.