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Question: The aqueous solution having \[pH{\text{ 11}}\] is how many times less basic the aqueous solution hav...

The aqueous solution having pH 11pH{\text{ 11}} is how many times less basic the aqueous solution having pH 8pH{\text{ }}8?
A.88
B.3030
C.300300
D.10001000

Explanation

Solution

The pH of a solution is a measure of hydrogen ion concentration, which in turn is a measure of its acidity. pH is calculated using the formula given below:
pH=−log⁡10[H+]pH = - {\log _{10}}[{H^ + }]
As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration.
pOH=−log⁡10[OH−]pOH = - {\log _{10}}[O{H^ - }]

Complete answer:
H2O⇄H++OH−{H_2}O \rightleftarrows {H^ + } + O{H^ - }
The equilibrium constant for this reaction, Kw{K_w} is the product of H+{H^ + } and OH−O{H^ - } concentrations. This relationship may be expressed as:
⇒Kw=[H+][OH−]\Rightarrow {K_w} = [{H^ + }][O{H^ - }]
At 25∘C{25^ \circ }C,
Kw=[H+][OH−]=10−14{K_w} = [{H^ + }][O{H^ - }] = {10^{ - 14}}
Using this information, we can now solve the problem
For aqueous solution having pH 11pH{\text{ 11}}
⇒pH=11\Rightarrow pH = 11
−log⁡[H+]=11- \log [{H^ + }] = 11
log⁡[H+]=−11\log [{H^ + }] = - 11
[H+]=antilog⁡(−11)[{H^ + }] = anti\log ( - 11)
⇒[H+]=10−11\Rightarrow [{H^ + }] = {10^{ - 11}}
We know that, [H+][OH−]=10−14[{H^ + }][O{H^ - }] = {10^{ - 14}}
[OH−]=10−14[H+]=10−1410−11[O{H^ - }] = \dfrac{{{{10}^{ - 14}}}}{{[{H^ + }]}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 11}}}}
⇒[OH−]=10−3\Rightarrow [O{H^ - }] = {10^{ - 3}}
Therefore, [OH−]1=10−3{[O{H^ - }]_1} = {10^{ - 3}}
For aqueous solution having pH 8pH{\text{ }}8
pH=8pH = 8
−log⁡[H+]=8- \log [{H^ + }] = 8
log⁡[H+]=−8\log [{H^ + }] = - 8
[H+]=antilog⁡(−8)[{H^ + }] = anti\log ( - 8)
⇒[H+]=10−8\Rightarrow [{H^ + }] = {10^{ - 8}}
[OH−]=10−14[H+]=10−1410−8[O{H^ - }] = \dfrac{{{{10}^{ - 14}}}}{{[{H^ + }]}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 8}}}}
⇒[OH−]=10−6\Rightarrow [O{H^ - }] = {10^{ - 6}}
Therefore, [OH−]2=10−6{[O{H^ - }]_2} = {10^{ - 6}}
On taking ratio of concentration of hydroxyl ion,
⇒[OH−]1[OH−]2=10−310−6=103\Rightarrow \dfrac{{{{[O{H^ - }]}_1}}}{{{{[O{H^ - }]}_2}}} = \dfrac{{{{10}^{ - 3}}}}{{{{10}^{ - 6}}}} = {10^3}
Hence, aqueous solution having pH 11pH{\text{ 11}}is 103{10^3}i.e. 10001000times more basic than solution having pH 8pH{\text{ }}8.

Option(D) is correct.

Note:
Acidic solutions (solutions with higher concentrations of H+{H^ + } ions) are measured to have lower pH values than basic or alkaline solutions. At 25∘C{25^ \circ }C, solutions with a pH less than 7 are acidic, and solutions with a pH greater than 7 are basic. Solutions with a pH of 7 at this temperature are neutral (e.g. pure water). The neutral value of the pH depends on the temperature, being lower than 7 if the temperature increases. The pH value can be less than 0 for very strong acids or greater than 14 for very strong bases.