school
Acemy

Acid-Base Titrations

AP Chemistry· difficulty 4/5

half-eq eq pt Volume NaOH (mL) pH

25.00 mL of 0.100 M acetic acid (Ka=1.8×10−5K_a = 1.8\times 10^{-5}) is titrated with 0.100 M NaOH. What is the pH at the equivalence point?

  • A

    pH≈5.27pH \approx 5.27

  • B

    pH=7.00pH = 7.00

  • C

    pH≈8.72pH \approx 8.72

    check_circle
  • D

    pH≈4.74pH \approx 4.74

Explanation

At eq pt, all acid converted to acetate; volume = 50.00 mL, [A−][A^-] = 0.0500 M. KbK_b = Kw/KaK_w/K_a = 5.56e-10. [OH−][OH^-] = (5.56e−10)(0.05)\sqrt{(5.56e-10)(0.05)} = 5.27e-6. pOH = 5.28, pH = 8.72.

Want 10 more like this — adaptive to your weak spots?

Practice 10 like thisarrow_forwardPredict my AP score

Related questions

auto_awesomeAll Toolsquestion_answerHomework Q&Amenu_bookAP & SAT PreptimelineLearn Mode