Question

e concentration of propionic acid in the solution? The K, for propionic acid is 1.3 x 105 (for HCaHsO2 from Table 16.1. 16.65 What is the pH of a 0.0 25 M aqueous solution of sodium propionate, NaC3HsO2? What is 4 5

Answer 1

Answer

pH = 8.64

concentration of propionic acid = 4.38×10^-6M

Explanation

C3H5O2^- ion is partly hydrolysed by water

C3H5O2^- ​​​​​​(aq) + H2O(l) <--------> HC3H5O2(aq) + OH^-(aq)

Kb = [HC3H5O2][OH^-]/[C3H5I2^-]

Kb = Kw /Ka

Kw = ionic product of water, 1.00×10^-14

Kb = 1.00×10^-14/1.3 ×10^-5 = 7.69×10^-10

Initial concentration

[C3H5O2^-] = 0.025

[HC3H5O2] = 0

[OH^-] = 0

change in conctration

[C3H5O2^-] = -x

[HC3H5O2] = +x

[OH^-] = +x

Equillibrium concentration

[C3H5O2^-]= 0.025 - x

[HC3H5O2] = x

[OH^-] = x

so,

x²/(0.025 - x) = 7.69×10^-10

we can assume , 0.025 - x = 0.025 because x is small value

x² /0.025 = 7.69×10^-10

x² = 1.92 ×10^-11

   x = 4.38 × 10^-6

Therefore

[OH^-] = 4.38×10^-6M

[HC3H5O2] = 4.38×10^-6M

pOH = -log[OH^-] = -log(4.38 ×10^-6) = 5.36

pH = 14 - pOH

pH = 14 - 5.36

pH = 8.64