
The pH value of 100 litre aqueous solution containing 4 gram of \[{\text{NaOH}}\]?
A.3
B.9
C.11
D.14
Answer
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Hint: The \[{\text{pH}}\] of a solution is defined as the negative logarithm to the base 10 of the hydrogen ion concentration is a solution. We shall calculate the moles of sodium hydroxide present and then its molarity. The molarity will be used to calculate the pH.
Formula Used: ${\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume of solution}}}}$
${\text{pH = }} - {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {{{\text{H}}^{\text{ + }}}} \right]$, where $\left[ {{{\text{H}}^{\text{ + }}}} \right]$ is the hydrogen ion concentration of the solution.
Complete step by step solution:
Since sodium hydroxide is a basic substance so we have to determine the \[{\text{pOH}}\] of the solution first since, \[{\text{pH + pOH }} = 14\]. Where, the \[{\text{pH}}\] is the negative logarithm to the base 10 hydroxyl ion concentration of the solution.
The molecular weight of sodium hydroxide = $23 + 16 + 1 = 40{\text{ g/mol}}$
Number of moles of 4 gram sodium hydroxide = $\dfrac{4}{{40}} = 0.1$ moles in 1 litre solution.
So in 100 litre the molarity will be = $\dfrac{{0.1}}{{100}} = 0.0001{\text{moles/ litre}}$.
The pOH of the solution = ${\text{pOH = }} - {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Hence, \[{\text{pOH}}\]=${\text{ - lo}}{{\text{g}}_{{\text{10}}}}\left[ {{{10}^{ - 3}}} \right] = 3$.
The \[{\text{pH}} = 14 - {\text{pOH}} = 14 - 3 = 11\]
Hence, the correct answer is option C.
Note: pH is a very important factor in different conditions. The pH of a reaction can even alter the rate of a reaction. The pH of water determines the solubility and the precipitation of different salts. The pH of a solution can range from 0 to 14. Where 0 is the value of for the most acidic solution and 14 is the value of the most basic solution. The pOH scale is just the opposite of the pH scale and here the value of 0 is most the most basic solution, while 14 is the value of the most acidic solution.
Formula Used: ${\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume of solution}}}}$
${\text{pH = }} - {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {{{\text{H}}^{\text{ + }}}} \right]$, where $\left[ {{{\text{H}}^{\text{ + }}}} \right]$ is the hydrogen ion concentration of the solution.
Complete step by step solution:
Since sodium hydroxide is a basic substance so we have to determine the \[{\text{pOH}}\] of the solution first since, \[{\text{pH + pOH }} = 14\]. Where, the \[{\text{pH}}\] is the negative logarithm to the base 10 hydroxyl ion concentration of the solution.
The molecular weight of sodium hydroxide = $23 + 16 + 1 = 40{\text{ g/mol}}$
Number of moles of 4 gram sodium hydroxide = $\dfrac{4}{{40}} = 0.1$ moles in 1 litre solution.
So in 100 litre the molarity will be = $\dfrac{{0.1}}{{100}} = 0.0001{\text{moles/ litre}}$.
The pOH of the solution = ${\text{pOH = }} - {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Hence, \[{\text{pOH}}\]=${\text{ - lo}}{{\text{g}}_{{\text{10}}}}\left[ {{{10}^{ - 3}}} \right] = 3$.
The \[{\text{pH}} = 14 - {\text{pOH}} = 14 - 3 = 11\]
Hence, the correct answer is option C.
Note: pH is a very important factor in different conditions. The pH of a reaction can even alter the rate of a reaction. The pH of water determines the solubility and the precipitation of different salts. The pH of a solution can range from 0 to 14. Where 0 is the value of for the most acidic solution and 14 is the value of the most basic solution. The pOH scale is just the opposite of the pH scale and here the value of 0 is most the most basic solution, while 14 is the value of the most acidic solution.
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