A solution with [H+] = 1.0e-4 M has pH and pOH values; calculate pH and pOH.

A solution with [H+] = 1.0e-4 M has pH and pOH values; calculate pH and pOH.

• A. pH = 10; pOH = 4
• B. pH = 3; pOH = 11
• C. pH = 7; pOH = 7
• D. pH = 4; pOH = 10

Answer: (D)

Understanding how pH and pOH relate to hydrogen and hydroxide ion concentrations is the key idea. pH is the negative base-10 logarithm of [H+], so with [H+] = 1.0×10^-4 M, pH = -log10(1.0×10^-4) = 4. For water at room temperature, Kw = [H+][OH-] ≈ 1.0×10^-14, so [OH-] = Kw/[H+] = 1.0×10^-14 / 1.0×10^-4 = 1.0×10^-10 M. Then pOH = -log10(1.0×10^-10) = 10. Since pH + pOH = 14 at this temperature, pOH = 14 − pH = 10. Therefore, the pH is 4 and the pOH is 10.

Understanding how pH and pOH relate to hydrogen and hydroxide ion concentrations is the key idea. pH is the negative base-10 logarithm of [H+], so with [H+] = 1.0×10^-4 M, pH = -log10(1.0×10^-4) = 4. For water at room temperature, Kw = [H+][OH-] ≈ 1.0×10^-14, so [OH-] = Kw/[H+] = 1.0×10^-14 / 1.0×10^-4 = 1.0×10^-10 M. Then pOH = -log10(1.0×10^-10) = 10. Since pH + pOH = 14 at this temperature, pOH = 14 − pH = 10. Therefore, the pH is 4 and the pOH is 10.