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A solution of $9 \%$ acid is to be diluted by adding $3 \%$ acid solution to it. The resulting mixture is to be more than $5 \%$ but less than $7 \%$ acid. If there is $460 L$ of the $9 \%$ solution, how many liters of $3 \%$ solution will have to be added?
Solution:
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Verified Answer
We have first $\mathrm{pH}$ value $=8.48$
and scond $\mathrm{pH}$ value $=8.35$
Let third $\mathrm{pH}$ value be $x$.
Since, average of the three $\mathrm{pH}$ value should lie between $8.2$ and $8.5$.
$$
\begin{aligned}
&\therefore 8.2 < \frac{8.48+8.35+x}{3} < 8.5 \\
&\Rightarrow 8.2 < \frac{16.83+x}{3} < 8.5 \\
&\Rightarrow 3 \times 8.2 < 16.83+x < 8.5 \times 3 \\
&\Rightarrow 24.6-16.83 < x < 25.5-16.83 \\
&\Rightarrow 7.77 < x < 8.67
\end{aligned}
$$
Thus, third $\mathrm{pH}$ value should lie between $7.77$ and $8.67$.
and scond $\mathrm{pH}$ value $=8.35$
Let third $\mathrm{pH}$ value be $x$.
Since, average of the three $\mathrm{pH}$ value should lie between $8.2$ and $8.5$.
$$
\begin{aligned}
&\therefore 8.2 < \frac{8.48+8.35+x}{3} < 8.5 \\
&\Rightarrow 8.2 < \frac{16.83+x}{3} < 8.5 \\
&\Rightarrow 3 \times 8.2 < 16.83+x < 8.5 \times 3 \\
&\Rightarrow 24.6-16.83 < x < 25.5-16.83 \\
&\Rightarrow 7.77 < x < 8.67
\end{aligned}
$$
Thus, third $\mathrm{pH}$ value should lie between $7.77$ and $8.67$.
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