Answer
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Hint: For this problem, firstly we will use the formula of pH, to find the concentration of hydrogen ion. After it, we will use ${{\text{M}}_{1}}{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}{{\text{V}}_{2}}$ to calculate the volume of water that is denoted by ${{\text{V}}_{2}}$ in the formula.
Complete step by step answer:
- In the given question, we have to calculate the volume of water by using the given data.
- As we know that pH is defined as the power of hydrogen which is used to measure the concentration of the hydrogen ion.
- pH is expressed as the negative of the logarithm of the concentration of the hydrogen ion or shown below:
$\text{pH = -log(}{{\text{H}}^{+}}\text{)}$
- So, now we can calculate the concentration of the hydrogen when the pH of the solution is 1 and 2 as shown:
At pH 1,
$1\text{ }\text{= -log(}{{\text{H}}^{+}}\text{)}$
${{\text{H}}^{+}}\text{ = 0}\text{.1M}$
And at pH 2,
$\text{2 }\text{= -log(}{{\text{H}}^{+}}\text{)}$
${{\text{H}}^{+}}\text{ = 0}\text{.01M}$
- Now, in the case of dilution, the formula that is used is:
${{\text{M}}_{1}}{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}{{\text{V}}_{2}}$ …. (1)
- Here, ${{\text{M}}_{1}}$ is the concentration of hydrogen & ${{\text{V}}_{1}}$ is the volume of water at pH 1 and ${{\text{M}}_{2}}$ is the concentration of hydrogen & ${{\text{V}}_{2}}$ is the volume of water at pH 2.
- So, putting all the value in equation (1) we will get:
$\text{0}\text{.1}\,\text{ }\times \text{ 1 = 0}\text{.01 }\times \text{ }{{\text{V}}_{2}}$
${{\text{V}}_{2}}\text{ = }\dfrac{\text{0}\text{.1}\,\text{ }\times \text{ 1}}{\text{0}\text{.01}}\text{ = 10L}$
- But here we know that in the solution 1 L of the solution with pH 1 is added so the amount of water added with pH 2 will be 9L.
So, the correct answer is “Option C”.
Note: If we want to analyse the pH of the solution practically then we usually use litmus paper. If the colour of the litmus paper becomes red then the solution is considered as acidic and vice versa.
Complete step by step answer:
- In the given question, we have to calculate the volume of water by using the given data.
- As we know that pH is defined as the power of hydrogen which is used to measure the concentration of the hydrogen ion.
- pH is expressed as the negative of the logarithm of the concentration of the hydrogen ion or shown below:
$\text{pH = -log(}{{\text{H}}^{+}}\text{)}$
- So, now we can calculate the concentration of the hydrogen when the pH of the solution is 1 and 2 as shown:
At pH 1,
$1\text{ }\text{= -log(}{{\text{H}}^{+}}\text{)}$
${{\text{H}}^{+}}\text{ = 0}\text{.1M}$
And at pH 2,
$\text{2 }\text{= -log(}{{\text{H}}^{+}}\text{)}$
${{\text{H}}^{+}}\text{ = 0}\text{.01M}$
- Now, in the case of dilution, the formula that is used is:
${{\text{M}}_{1}}{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}{{\text{V}}_{2}}$ …. (1)
- Here, ${{\text{M}}_{1}}$ is the concentration of hydrogen & ${{\text{V}}_{1}}$ is the volume of water at pH 1 and ${{\text{M}}_{2}}$ is the concentration of hydrogen & ${{\text{V}}_{2}}$ is the volume of water at pH 2.
- So, putting all the value in equation (1) we will get:
$\text{0}\text{.1}\,\text{ }\times \text{ 1 = 0}\text{.01 }\times \text{ }{{\text{V}}_{2}}$
${{\text{V}}_{2}}\text{ = }\dfrac{\text{0}\text{.1}\,\text{ }\times \text{ 1}}{\text{0}\text{.01}}\text{ = 10L}$
- But here we know that in the solution 1 L of the solution with pH 1 is added so the amount of water added with pH 2 will be 9L.
So, the correct answer is “Option C”.
Note: If we want to analyse the pH of the solution practically then we usually use litmus paper. If the colour of the litmus paper becomes red then the solution is considered as acidic and vice versa.
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