Question

A 0.5-L unbuffered solution needs the pH adjusted from 4 to 3.5. How many microliters of a 2 molar HCl solution need to be added to adjust the pH?

Report your answer to four decimal places.

Answer 1

We know that concentration of H⁺ = 10^-pH

Initial concentration of H⁺ = 10^-4 M

Final concentration of H⁺ = 10^-3.5 = 3.16 * 10^-4 M

Additional concentration of H⁺ = final - initial = 3.16 * 10^-4 - 1 * 10^-4 = 2.16 * 10^-4 M

moles of H⁺ needed = Volume * molarity = 0.5 * 2.16 * 10^-4 = 1.08 * 10^-4 moles

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

V * 2 = 1.08 * 10^-4

SOlving, V = 5.4 * 10^-5 Liters = 54 microliters