# Sulfuric Acid with a molarity of .001 has a pH of 2.75. You want to dilute 1mL of this acid to a pH of 5.2. If you want to dilute 1mL of the original solution, what would the final volume of solution be?

May 19, 2018

282.1 ml

#### Explanation:

For the original solution:

$\textsf{p H = - \log \left[{H}^{+}\right] = 2.75}$

$\textsf{\left[{H}^{+}\right] = 0.00178 \textcolor{w h i t e}{x} \text{mol/l}}$

For the target solution:

$\textsf{p H = - \log \left[{H}^{+}\right] = 5.2}$

$\textsf{\left[{H}^{+}\right] = 6.31 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

This means the dilution factor to go from pH 2.75 to pH 5.2 will be:

$\textsf{\frac{0.00178}{6.31 \times {10}^{- 6}} = 282.1}$

This means you take 1 ml of the original solution and make it up to a final volume of 282.1 ml