Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.60.

Answer 1

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