Question

If $K_a=1.8\times {10}^{-5}$ for acetic acid, its degree of dissociation for 0.05 M solution would be:

• A. 0.0019
• B. 0.019
• C. 0.19
• D. 0.45

Answer 1

The correct option is B 0.019

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{\ominus }+H^{⨁}0.05000.05(1-\alpha )0.05\alpha 0.05\alpha K_a=\frac{\left[C{H}_{3}CO{O}^{\ominus }\right]\left[H^{⨁}\right]}{\left[C{H}_{3}COOH\right]}1.8\times {10}^{-5}=\frac{0.05{\alpha }^2}{0.051-\alpha }=\frac{0.05{\alpha }^2}{1-\alpha }(1-\alpha \approx 1)1.8\times {10}^{-5}=0.05{\alpha }^2\left(or\right)\alpha =\sqrt{\frac{1.8\times {10}^{-5}}{0.05}}=0.0189$