Answer
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Hint: \[pH\] is negative logarithm of concentration of \[{H^ + }\] and after finding concentration of \[{H^ + }\] just compare the two \[pH\] and concentration and it will give the answer easily by this simple calculation.
Step by step explanation: -
Before proceeding directly to the answer and its calculation part, first let us know about pH and its formula
So, \[pH\] is used to measure the acidity or basicity of a given solution and \[pH\] is defined also as a negative logarithm of concentration of Hydrogen ions. So numerically \[pH{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]
It is the amount of measure of free hydrogen ion present in the solution. It has a general range from 0 to 14 in which less pH means acidic solution and pH greater than 7 becomes basic or alkaline solution.
And \[pOH\] will be a negative logarithm of concentration of \[O{H^ - }\] ion.
Also \[pH{\text{ }} + {\text{ }}pOH{\text{ }} = {\text{ }}14\]
Now coming to the question, one solution has \[pH\] of 2 and the other solution has \[pH\] of 6.
So, solution 1 = \[pH{\text{ }} = {\text{ }}2\]
So \[2{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]
\[\left[ {{H^ + }} \right]{\text{ }} = {\text{ }}{10^{ - 2}}M\]
For second solution \[pH{\text{ }} = {\text{ }}6\]
So \[6{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]
\[\left[ {{H^ + }} \right]{\text{ }} = {\text{ }}{10^{ - 6}}M\]
So, comparing the concentration of \[{2^{nd}}\] solution and \[{1^{st}}\]solution.
$\frac{{{1^{st}}}}{{{2^{nd}}}} = \frac{{{{10}^{ - 2}}}}{{{{10}^{ - 6}}}} = {10^4}$
So, the 1st (pH=2) solution is \[{10^4}\] times more acidic than the 2nd (pH =6) solution.
Therefore, among the given options D is the correct one i.e. \[{10^4}\]
Note: To compare acidic or basic character in such type of similar questions, just look for the pH and pOH value and also if pH is less it means the solution is more acidic and vice versa. Some of the examples of acidic solutions are \[HCl, {\text{ }}HN{O_3}\] etc. and of the basic solution is \[NaOH\]. Water is considered to be neutral and has pH of 7.
Step by step explanation: -
Before proceeding directly to the answer and its calculation part, first let us know about pH and its formula
So, \[pH\] is used to measure the acidity or basicity of a given solution and \[pH\] is defined also as a negative logarithm of concentration of Hydrogen ions. So numerically \[pH{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]
It is the amount of measure of free hydrogen ion present in the solution. It has a general range from 0 to 14 in which less pH means acidic solution and pH greater than 7 becomes basic or alkaline solution.
And \[pOH\] will be a negative logarithm of concentration of \[O{H^ - }\] ion.
Also \[pH{\text{ }} + {\text{ }}pOH{\text{ }} = {\text{ }}14\]
Now coming to the question, one solution has \[pH\] of 2 and the other solution has \[pH\] of 6.
So, solution 1 = \[pH{\text{ }} = {\text{ }}2\]
So \[2{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]
\[\left[ {{H^ + }} \right]{\text{ }} = {\text{ }}{10^{ - 2}}M\]
For second solution \[pH{\text{ }} = {\text{ }}6\]
So \[6{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]
\[\left[ {{H^ + }} \right]{\text{ }} = {\text{ }}{10^{ - 6}}M\]
So, comparing the concentration of \[{2^{nd}}\] solution and \[{1^{st}}\]solution.
$\frac{{{1^{st}}}}{{{2^{nd}}}} = \frac{{{{10}^{ - 2}}}}{{{{10}^{ - 6}}}} = {10^4}$
So, the 1st (pH=2) solution is \[{10^4}\] times more acidic than the 2nd (pH =6) solution.
Therefore, among the given options D is the correct one i.e. \[{10^4}\]
Note: To compare acidic or basic character in such type of similar questions, just look for the pH and pOH value and also if pH is less it means the solution is more acidic and vice versa. Some of the examples of acidic solutions are \[HCl, {\text{ }}HN{O_3}\] etc. and of the basic solution is \[NaOH\]. Water is considered to be neutral and has pH of 7.
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