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Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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What is the pH of a 0.001 M NaOH solution?
3
11
10
4
11
A 0.001 M NaOH solution fully dissociates to give [OH⁻] = 1×10⁻³ M; the pOH is 3 and pH = 14 − 3 = 11.
A 0.001 M NaOH solution fully dissociates to give [OH⁻] = 1×10⁻³ M; the pOH is 3 and pH = 14 − 3 = 11.
Think of a bathtub full of water (neutral pH 7); adding a small amount of a strong base is like pouring a few drops of soap that quickly make the water alkaline, raising the pH noticeably.
Strong Base → Complete Dissociation → pOH = -log[OH⁻] → pH = 14 - pOH.
pOH=−log10[OH−] — defines pOH from hydroxide concentration
pH=14−pOH — relationship between pH and pOH at 25 °C
NaOH is a strong base that ionises completely in water: NaOH → Na⁺ + OH⁻. The hydroxide ion concentration determines pOH via pOH = -log[OH⁻]; pH is then obtained from the water ion‑product relation pH + pOH = 14 at 25 °C.
NaOH is a strong base and dissociates completely in aqueous solution.
At 25 °C, the product of [H⁺] and [OH⁻] is Kw=1×10−14, giving pH + pOH = 14.
pH calculation is essential in titration and buffer preparation.
Understanding strong base behavior helps in wastewater treatment.
Option A (pH 3) would correspond to a strong acid of 0.001 M, not a base.
Option C (pH 10) would arise from a 0.0001 M NaOH solution.
Option D (pH 4) is far too acidic for any NaOH concentration.
B is correct — a 0.001 M NaOH solution has a pH of 11.
Remember that for any strong acid or base, pH = –log[H⁺] or pH = 14 + log[OH⁻]; this shortcut avoids calculating pOH first.