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Question: Calculate pOH value for 0.025M \(C{H_3}COOH\) solution if dissociation constant for \(C{H_3}COOH\) i...

Calculate pOH value for 0.025M CH3COOHC{H_3}COOH solution if dissociation constant for CH3COOHC{H_3}COOH if 10−5{10^{ - 5}} m.

Explanation

Solution

This question is based on chemical equilibrium and we have to calculate pOH of the given acid in the given conditions. This can be done by writing the chemical equation of dissociation of the given acid and then by writing the equation at its equilibrium state.

Complete answer:
There are so many terms in the question which you might not be aware of. Let's talk about them one by one.
pOH is a measure of hydroxide ion concentration in a given solution. It is used to express the alkalinity of the solution. Aqueous solutions at 25oC with pOH less than 7 are alkaline in nature, pOH greater than 7 are acidic and pOH equal to 7 are neutral in nature. Hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional to the ion product of water. The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions, acidic if it contains a greater concentration of hydronium ions and basic if it contains a lesser concentration of hydronium ions than hydroxide ions. Relation between pOH and concentration of OH- ion is denoted by:
pOH=−log⁡[OH−]pOH = - \log [O{H^ - }]
Another term is dissociation constant. In chemistry a dissociation constant is a specific type of equilibrium constant that measures the propensity of a larger object to dissociate reversibly into smaller components, as when a molecule falls apart into its component molecules, or when a salt splits up into its component ions. In the special case of salts, the dissociation constant can also be called an ionization constant.
Now, to find the answer, let's first write the concentration of both reactant and product at equilibrium.
So, we can write,
K=[CH3COO−][H+][CH3COOH]K = \dfrac{{[C{H_3}CO{O^ - }][{H^ + }]}}{{[C{H_3}COOH]}}
Substituting the values, we get
10−5=x20.025{10^{ - 5}} = \dfrac{{{x^2}}}{{0.025}}
By solving, we will get x = 5×10−45 \times {10^{ - 4}}
Now,
x = [H+{H^ + }] = 5×10−45 \times {10^{ - 4}}
pH=−log⁡(5×10−4)=3.3 pOH=14−pH=14−3.3=10.7  pH = - \log (5 \times {10^{ - 4}}) = 3.3 \\\ pOH = 14 - pH = 14 - 3.3 = 10.7 \\\
Note:
Remember that the sum of pH and pOH is equal to 14 only at room temperature. Its values may change according to the temperature given since the dissociation constant gets changed. But if nothing is not mentioned in the question then take their sum as 14 otherwise it will be mentioned.