1.The degree of ionization of 0.4 M acetic acid will be (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}Ka=1.8×10−5)a.6.71×10−36.71 \times 10^{-3}6.71×10−3b.1.6×10−31.6 \times 10^{-3}1.6×10−3c.0.4×1.8×10−50.4 \times 1.8 \times 10^{-5}0.4×1.8×10−5d.1.8×10−51.8 \times 10^{-5}1.8×10−5Login to continueOnly logged in users canattempt or see the solution.
