Question

Jackie has two solutions that are 2 percent sulphuric acid and 12 percent sulphuric acid by volume respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of solution that is 5 percent sulphuric acid, approximately how many liters of 2 percent solution will be present.

Answer 1

Hint:Consider k liters of the 2% sulphuric acid and (60-k) liters of 12% sulphuric acid is present in 60-liter final solution.
Then find the amount of sulphuric acid in 5% solution by using the formula: \[Amount\text{ }of\text{ }sulphuric~acid\text{ }by\text{ }volume=\left( Volume\text{ }of\text{ }the\text{ }solution \right)\left( \dfrac{percentage\text{ }of\text{ }sulphuric\text{ }acid}{100} \right)\]
Equate it to the sum of the amounts of sulphuric acid present in 2% and 12% solution to get a linear equation in k. Solve the equation to get the value of k.

Complete step-by-step answer:
Let’s say : 2% of sulphuric acid solution by volume as \[{{s}_{1}}\] solution. 12% of sulphuric acid solution by volume as \[{{s}_{2}}\] solution.
Therefore, the final solution has k liters of \[{{s}_{1}}\] solution and (60-k) liters of \[{{s}_{2}}\] solution. In the question it is given that 5% of sulphuric acid by volume is present in the final solution.
\[\Rightarrow Amount\text{ }of\text{ }sulphuric\text{ }acid\text{ }present\text{ }in\text{ }the\text{ }final\text{ }solution=\left( Volume\text{ }of\text{ }the\text{ }solution \right)\left( \dfrac{percentage\text{ }of\text{ }sulphuric\text{ }acid}{100} \right)\]
\[=60\left( \dfrac{5}{100} \right)\]
\[=60\left( \dfrac{1}{20} \right)\]
$\Rightarrow $ Amount of the sulphuric acid present in the final solution = 3 liters \[.......\left( 1 \right)\]
Now, we will find the amount of sulphuric acid present in ${{s}_{1}}$ solution.
Amount of sulphuric acid by solution \[{{s}_{1}}\] in volume = \[\left( Volume\,of\,the\,{{s}_{1}}\,solution \right)\left( \dfrac{\text{percentage of sulphuric acid in }{{\text{s}}_{1}}\text{ solution }}{100} \right)\]
\[=k\left( \dfrac{2}{100} \right)\]
\[=\dfrac{2k}{100}\]
$\Rightarrow $ Amount of sulphuric acid present in the \[{{s}_{1}}\] solution\[=\dfrac{2k}{100}\]\[.......\left( 2 \right)\]
Now, we will find the amount of sulphuric acid in \[{{s}_{2}}\] solution.
Amount of sulphuric acid by volume in \[{{s}_{2}}\] solution = \[\left( Volume\text{ }of\text{ }the\text{ }{{\text{s}}_{2}}\text{ }solution \right)\left( \dfrac{\text{percentage of sulphuric acid in }{{\text{s}}_{2}}\text{ solution }}{100} \right)\]
\[=(60-k)\left( \dfrac{12}{100} \right)\]
\[=\dfrac{12\left( 60-k \right)}{100}\]
$\Rightarrow $ Amount of sulphuric acid present in the \[{{s}_{1}}\] solution \[=(60-k)\left( \dfrac{12}{100} \right)\] \[.......\left( 3 \right)\]
Now, we know that the amount of sulphuric acid present in 5% solution is equal to the sum of the amounts of sulphuric acid present in 2% and 12% solution, i.e. Amount of sulphuric acid in the final solution by volume = amount of sulphuric acid by solution \[{{s}_{1}}\] in volume + amount of sulphuric acid by solution \[{{s}_{2}}\] in volume. \[.......\left( 4 \right)\]
Now, we will substitute equation 1,2,3 in equation 4. On substitution, we get:
\[3=\dfrac{2k}{100}+12\left( \dfrac{60-k}{100} \right)\]
\[\Rightarrow 3=\dfrac{720-10k}{100}\]
\[\Rightarrow 300=720-10k\]
$\Rightarrow 420=10k$

\[\Rightarrow k=42\]
Therefore, 42 liters of the 2% sulphuric acid by volume is contained in the final solution.

Note: We can equate the amounts before and after mixing only if the given two solutions do not react with each other. If they react with each other we have to consider the reaction and calculate accordingly.Students should use the formula to find the amount of sulphuric acid in 5% solution: \[Amount\text{ }of\text{ }sulphuric~acid\text{ }by\text{ }volume=\left( Volume\text{ }of\text{ }the\text{ }solution \right)\left( \dfrac{percentage\text{ }of\text{ }sulphuric\text{ }acid}{100} \right)\] to solve this question.