Answer:
15.1 mL of the HCl solution should be added
Explanation:
First off, let's use the definition of pH to find the desired number of H⁺ moles:
pH = -log[H⁺]
2.80 = -log[H⁺]
= [H⁺]
[H⁺] = 1.58 x 10⁻³ M
Then we use the definition of Molarity [C=n/V]:
[H⁺] = 1.58 x 10⁻³ M = molesH⁺ / 1.00 L
molesH⁺ = 1.58 x 10⁻³
So in the end there needs to be 1.58 x10⁻³ moles of H⁺ to achieve the desired pH.
Now let's calculate the added amounts of HCl and NaOH
mL HCl = 100 - 84 = 16 mL = 0.016 L
mL NaOH = 100 - 88 = 12 mL = 0.012 L
mol HCl = 0.016 L * 7.00x10⁻² M = 1.12x10⁻³ mol HCl = mol H⁺
mol HCl = 0.012 L * 5.00x10⁻² M = 6.0x10⁻⁴ mol NaOH = mol OH⁻
OH⁻ reacts with H⁺, producing water. So the moles of H⁺ remaining in the solution are:
1.12x10⁻³ - 6.0x10⁻⁴ = 5.2x10⁻⁴ mol H⁺
This means that the moles of H⁺ that we need to add are:
1.58x10⁻³ - 5.2x10⁻⁴ = 1.06x10⁻³ mol H⁺
Finally we calculate the required volume of HCl:
C = n/V
7.00 x10⁻² M = 1.06x10⁻³ mol H⁺ / V
V = 0.0151 L = 15.1 mL
