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Question: Calculate \[\left[ {{H}^{+}} \right]~\] and \[\left[ CHC{{l}_{2}}CO{{O}^{-}} \right]~\] in a solutio...

Calculate [H+] \left[ {{H}^{+}} \right]~ and [CHCl2COO−] \left[ CHC{{l}_{2}}CO{{O}^{-}} \right]~ in a solution that is 0.01M(HCl)0.01M\left( HCl \right) and 0.01M(CHCl2COOH)0.01M(CHC{{l}_{2}}COOH). Take (Ka=2.55×10−2).\left( {{K}_{a}}=2.55\times {{10}^{-2}} \right).

Explanation

Solution

To answer this question, we must know that pH stands for power of hydrogen. It is actually based on the concentration of hydrogen ions in an aqueous solution.

Complete step by step solution:
We know that pH is a measure of how acidic or basic a substance is. The scale ranges from 0−140-14 with seven being the neutral mark. A pH of less than seven associated with each pH. Strong acids and bases are compounds that are completely dissociated in water. Under normal circumstances this means that the concentration of hydrogen ions in acidic solution can be taken to be equal to the concentration of the acid. The pH is then equal to minus the logarithm of the concentration value. Hydrochloric acid is a strong acid which makes our calculation easier.
pH is essential to determine the quality of water. It also indicates its solubility, biological availability of chemical constituents and heavy metals which may render it undrinkable.
[HCl]=0.01M,[CHCl2COOH]=0.01M\left[ HCl \right]=0.01M,\left[ CHC{{l}_{2}}COOH \right]=0.01M and Ka=2.55×10−2{{K}_{a}}=2.55\times {{10}^{-2}}
The reaction is given by HCl→H++Cl−HCl\to {{H}^{+}}+C{{l}^{-}} which can rewritten as 0.01M→0.01+0.010.01M\to 0.01+0.01
Thus the reaction; CHCl2COOH⟷CHCl2COO−+H+CHCl2COOH\underset{{}}{\overset{{}}{\longleftrightarrow}}CHC{{l}_{2}}CO{{O}^{-}}+{{H}^{+}}
Here at T=0 0.01M\text{T=0 0.01M}
At T=t 0.01-x x x+0.01\text{T=t 0.01-x x x+0.01}
For Ka=x(x+0.01)(0.01−x)=0.0255{{K}_{a}}=\dfrac{x\left( x+0.01 \right)}{\left( 0.01-x \right)}=0.0255
Now we have to find value of xx
x2+0.01x=0.0255(0.01−x){{x}^{2}}+0.01x=0.0255(0.01-x)
The above equation will give as quadratic eqn
x2+0.0355x−0.000255=0{{x}^{2}}+0.0355x-0.000255=0

⇒x=−0.0355±(0.0355)2+4×0.0002552\Rightarrow x=\dfrac{-0.0355\pm \sqrt{{{\left( 0.0355 \right)}^{2}}+4\times 0.000255}}{2}
Thus the value of x we get is;
x=0.01222−0.006x=\dfrac{0.0122}{2}-0.006
Therefore, [H+]=0.01+0.006=0.016\left[ {{H}^{+}} \right]=0.01+0.006=0.016 and [CHCl2COOH]=x=0.006\left[ CHC{{l}_{2}}COOH \right]=x=0.006

Note: We must always remember that as concentration changes with temperature, the pH scale also shifts. With increase in temperature the pH decreases. This however does not mean that water becomes acidic at higher temperature. It simply means that the scale shifts and seven is the neutral point only at normal temperature.