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Question

Answers

(A) 3.699

(B) 10.301

(C) 4.691

(D) 9.301

Answer

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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 \[N{H_4}OH\] solution. It has been shown that \[N{H_4}OH\] is present in millimolar concentration.

\[N{H_4}OH = 0.001M\]……. (1)

The ammonium hydroxide is a weak base, so \[N{H_4}OH\] will dissociate as \[O{H^ - }\] ions partially.

The degree of dissociation of \[N{H_4}OH\] is 20 %

degree of dissociation \[ = 20\% = \dfrac{{20}}{{100}} = 0.2\]……. (2)

From the above value concentration of \[O{H^ - }\]ions can be calculated

\[{\text{Concentration of O}}{{\text{H}}^{\text{ - }}}{\text{ = Concentration of solution}} \times {\text{degree of dissociation}}\]

\[[O{H^ - }] = 0.001 \times 0.2 = 2 \times {10^{ - 4}}\]…… (3)

We know that water has pOH = 7

Therefore, the concentration of \[O{H^ - }\] ions will be

\[pOH = - \log [O{H^ - }]\]……... (4)

\[7 = - \log [O{H^ - }]\]

\[[O{H^ - }] = {10^{ - 7}}\]….…... (5)

- To find the total concentration of \[O{H^ - }\] we should add equation (3) and (5)

\[total [O{H^ - }] = (2 \times {10^{ - 4}}) + {10^{ - 7}}\]

\[total [O{H^ - }] = {10^{ - 4}}(2 + 0.001)\]

\[total [O{H^ - }] = 2.001 \times {10^{ - 4}}\]

- Substituting \[[O{H^ - }]\] in equation (4)

\[pOH = - \log (2.001 \times {10^{ - 4}})\]

\[pOH = 3.6988\]

pH can be calculated using the equation

\[pH = 14 - pOH\]…… (6)

\[pH = 14 - 3.6988\]

\[pH = 10.301\]

The pH of a millimolar solution of ammonium hydroxide which is 20 % dissociated is 10.301.

Acids and bases are measured using pH or pOH scale. This scale provides the measure of \[{H^ + }\] or \[O{H^ - }\] 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.