Question

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook).
Part A. 0.270 M

Part B. 7.84×10-2 M

Part C. 1.97×10-2 M

Answer 1

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$= $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608  × 100%

= 2.60%