What is the ph of 1.0 l of water to which (a) 1.5 ml of 3.0 m hcl or (b) 1.5 ml of 3.0 m naoh has been added?

(a) 2.3

(b) 11.7

Explanation:

(a)

The moles of HCl in 1.5 mL of 3.0 M HCl are:

1.5 × 10⁻³ L × 3.0 mol/L = 4.5 × 10⁻³ mol

When 1.5 × 10⁻³ L of this solution are added to 1.0 L of water, the volume of the solution is:

1.0 L + 1.5 × 10⁻³ L = 1.0015 L

The concentration of HCl in this solution is:

4.5 × 10⁻³ mol/1.0015 L = 4.5 × 10⁻³ M

HCl is a strong monoprotic acid, so [H⁺] = 4.5 × 10⁻³ M. The pH is:

pH = -log [H⁺] = -log 4.5 × 10⁻³ = 2.3

(b)

The moles of NaOH in 1.5 mL of 3.0 M NaOH are:

1.5 × 10⁻³ L × 3.0 mol/L = 4.5 × 10⁻³ mol

When 1.5 × 10⁻³ L of this solution are added to 1.0 L of water, the volume of the solution is:

1.0 L + 1.5 × 10⁻³ L = 1.0015 L

The concentration of NaOH in this solution is:

4.5 × 10⁻³ mol/1.0015 L = 4.5 × 10⁻³ M

NaOH is a strong monohydroxide, so [OH⁻] = 4.5 × 10⁻³ M. The pOH is:

pOH = -log [OH⁻] = -log 4.5 × 10⁻³ = 2.3

The pH is

pH + pOH = 14.0

pH = 14.0 - pOH = 14.0 - 2.3 = 11.7

the answer is symmetrical : )