Question

A 0.5-L unbuffered solution needs the pH adjusted from 6 to 3. How many microliters of a 2 molar HCl solution need to be added to adjust the pH?

Report your answer to four decimal places.

For this question:

• Assume the volume of HCl being added is negligible and the total volume of the solution does not change.
• The volume of the solution is *not* 1 L like it was in the practice pools.
• The question requests your answer be reported in μL (microliters)!!

Answer 1

We know that: pH = -log([H⁺])

From this: [H⁺] = 10^-pH

Initial [H⁺] = 10^-6 M

Final [H⁺] = 10^-3 M

Change in [H⁺] = 10^-3 - 10^-6 = 9.99 * 10^-5 M

Moles of H⁺ = [H⁺] * Volume of solution = 9.99 * 10^-5 * 0.5 = 5 * 10^-5 moles

Molarity of HCl * Volume of HCl = moles of H⁺

2 * V = 5 * 10^-5

Solving, V = 2.5 * 10^-5 = 25 microliters