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Weak Acid and Base Equilibria

AP Chemistry· difficulty 4/5

What is the pH of 0.250 M HC2H3O2HC_2H_3O_2 (Ka=1.8×10−5K_a = 1.8\times 10^{-5})?

  • A

    pH≈4.74pH \approx 4.74

  • B

    pH≈3.40pH \approx 3.40

  • C

    pH≈2.67pH \approx 2.67

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  • D

    pH≈5.30pH \approx 5.30

Explanation

[H+]=KaC[H^+] = \sqrt{K_a C} = (1.8×10−5)(0.250)\sqrt{(1.8\times 10^{-5})(0.250)} = 4.5×10−6\sqrt{4.5\times 10^{-6}} = 2.12×10−3\times 10^{-3}. pH = −log⁡-\log(2.12e-3) = 2.67.

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