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Question: The value \({{K}_{a}}\) of \({{C}_{6}}{{H}_{5}}OH\) is \(1.0\times {{10}^{-10}}\) what is the \(pH\)...

The value Ka{{K}_{a}} of C6H5OH{{C}_{6}}{{H}_{5}}OH is 1.0×10−101.0\times {{10}^{-10}} what is the pHpH of a 0.1M C6H5O−0.1M\text{ }{{\text{C}}_{6}}{{H}_{5}}{{O}^{-}} solution?
A. 10.5110.51
B. 11.0411.04
C. 11.5011.50
D. 1212

Explanation

Solution

Since we need to find the pHpH of the anion, first find the pOHpOH of the ion by using the relation pOH=kbCpOH=\sqrt{{{k}_{b}}C} and then find the value by the relation pH=14−pOHpH=14-pOH. Using the above relations or the formulas, find the required pHpH of the anion.

Complete step by step solution:
Given the value of Ka{{K}_{a}} is equal to 1.0×10−101.0\times {{10}^{-10}}
We know that kb=kwka{{k}_{b}}=\dfrac{{{k}_{w}}}{{{k}_{a}}} and the value of kw{{k}_{w}} is equal to 1.0×10−141.0\times {{10}^{-14}}
By putting the above two values in the above relation, we obtain
kb=kwka=1.0×10−141.0×10−10=1.0×10−4{{k}_{b}}=\dfrac{{{k}_{w}}}{{{k}_{a}}}=\dfrac{1.0\times {{10}^{-14}}}{1.0\times {{10}^{-10}}}=1.0\times {{10}^{-4}}
Given that the concentration of the acid C6H5OH{{C}_{6}}{{H}_{5}}OH is 0.1M 0.1M\text{ }
We know that for a base phenoxide ion, the expression for the pOHpOH of the solution is given by the expression
pOH=−log⁡kbCpOH=-\log \sqrt{{{k}_{b}}C}
Put the above values in this formula and find out the value of the pOHpOH
pOH=−log⁡kbC=−log⁡(1.0×10−4×0.1)=2.5pOH=-\log \sqrt{{{k}_{b}}C}=-\log \sqrt{\left( 1.0\times {{10}^{-4}}\times 0.1 \right)}=2.5
Hence we find the value of the pOHpOH of the phenoxide ion is 2.5
We know the relation between the pHpH & pOHpOH as
pH=14−pOHpH=14-pOH
Put the above pOHpOH value in the formula and find the required pHpH of the anion
After substituting the values, we obtain
pH=14−pOH=14−2.5=11.5pH=14-pOH=14-2.5=11.5
Therefore the required value of thepHpH of the phenoxide ion is obtained as
pH=11.5pH=11.5

Hence option (C) is the correct answer.

Note: For a cation we can directly find the value of the pHpH by using a certain relation we are having where as for an anion we cannot find the pHpH value directly. We are having the formula as pH=−log⁡[H+]pH=-\log \left[ {{H}^{+}} \right], with the cation but not with the anion. But we can find the value of pOHpOH for an anion or a base and using this pOHpOH we need to find the value of pHpH using the relation pH=14−pOHpH=14-pOH. This is due to the fact that the acid or a cation is having the less pHpH value whereas for the anion it is having large pHpH value.