Answer:
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) =
1661μL of a 6M HCl you need to add
