What is the pH of 500 mL of water after the addition of 1 mL of 2M NaOH?

pH = 11.60

Explanation:

When we add a base, we are increasing [OH⁻], so the pH will be totally basic.

pH of water = 7

Basic pH > 7

We are adding 2 M . 1 mL = 2 mmoles

2 mmoles of OH⁻ are contained in 501 mL of total volume.

[OH⁻] = 2 mmol / 501 mL =  3.99×10⁻³ M

- log 3.99×10⁻³ M = 2.39 → pOH

pH = 14 - pOH → 11.61

Water equilibrium

2H₂O  ⇄  H₃O⁺   +  OH⁻          Kw = 1×10⁻¹⁴

NaOH  →  Na⁺   +  OH⁻

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