A 0.1-L unbuffered solution needs the pH adjusted from 3.5 to 1. How many microliters of a 6 molar HCl solution need to be added to adjust the pH

1661μL of a 6M HCl you need to add

Explanation:

pH is defined as -log[H⁺] ([H⁺] =10^{-pH}), the initial and final concentrations of [H⁺] you need are:

Initial [H⁺] = 10^{-3.5} = 3.16x10⁻⁴M H⁺

Final [H⁺] = 10^{-1} = 0.1M H⁺

In moles, knowing volume of the solution is 0.1L:

Initial [H⁺] = 0.1L ₓ (3.16x10⁻⁴mol H⁺ / L) = 3.16x10⁻⁵moles H⁺

Final [H⁺] = 0.1L ₓ (0.1mol H⁺ / L) = 0.01 moles H⁺.

That means, moles of H⁺ you need to add to the solution is:

0.01mol - 3.16x10⁻⁵moles = 9.9684x10⁻³ moles of H⁺.

A solution of HCl dissociates in H⁺ and Cl⁻ ions, that means moles of HCl added are equal to moles of H⁺. As you need to add 9.9684x10⁻³ moles of H⁺ = 9.9684x10⁻³ moles of HCl:

9.9684x10⁻³ moles of HCl ₓ (1L / 6mol) = 1.6614x10⁻³L

In μL:

1.661x10⁻³L × (1x10⁶μL / 1L) =

pv=nrt

explanation: