Given that for ethanoic acid pKa=4.75, calculate the pH of sodium ethanoate 0.1M.
C2H50Na + H2O ==> Na^+ + C2H5O^-
The ethanoate ion hydrolyzes as follows:
C2H5O^- + HOH --> C2H5OH + OH^-
Khydrolysis = Kb = Kw/Ka = (C2H5OH)(OH^-)/(C2H5O^-).
(C2H5OH) = y
(OH^-) = y
(C2H5O^-) = 0.1 - y
solve for y and convert to pOH and from there to pH.
Check my thinking. Post your wor if you get stuck.
To calculate the pH of a sodium ethanoate solution, we need to consider the dissociation of sodium ethanoate in water.
Sodium ethanoate (CH3COONa) is a salt that dissociates into sodium ions (Na+) and ethanoate ions (CH3COO-) in water. Since Na+ ions do not affect the pH of a solution, we only need to consider the ethanoate ions.
The ethanoate ion (CH3COO-) can react with water to produce ethanoic acid (CH3COOH) and hydroxide ions (OH-). This reaction can be represented as:
CH3COO- + H2O ⇌ CH3COOH + OH-
The equilibrium constant for this reaction is given by the dissociation constant (Ka) of the ethanoic acid. The pKa of ethanoic acid is given as 4.75, which means that:
pKa = -log10(Ka)
4.75 = -log10(Ka)
Now, we can find the value of Ka by taking the antilog of both sides:
Ka = 10^(-4.75)
Ka = 1.78 * 10^(-5)
Since sodium ethanoate is a strong electrolyte, it ionizes completely in water, meaning that the initial concentration of ethanoate ions is equal to the concentration of sodium ethanoate (0.1 M).
Now, we can set up an ICE (initial, change, equilibrium) table to determine the concentration of ethanoate ions at equilibrium:
CH3COO- + H2O ⇌ CH3COOH + OH-
Initial 0.1 M 0 M 0 M
Change -x M +x M
Equilibrium x M x M x M
Using the equilibrium constant expression, we can write:
Ka = [CH3COOH][OH-]/[CH3COO-]
Since the concentration of water remains nearly constant during the reaction, we can assume that [H2O] = 1 (because pure water is taken as 1 in the expression). Therefore, we can simplify the equation:
Ka = [CH3COOH][OH-] / [CH3COO-]
1.78 * 10^(-5) = x * x / 0.1
Solving for x, we find:
x = √[(1.78 * 10^(-5)) * 0.1]
x ≈ 0.0027 M
Now that we have the concentration of hydroxide ions (OH-) at equilibrium, we can use the fact that the concentration of hydrogen ions (H+) in water is equal to the concentration of hydroxide ions to calculate the pH. The concentration of H+ ions can be calculated using the equation:
pH = -log10 [H+]
Since the concentration of H+ ions in water is 10^(-14) M, we can substitute the concentration of OH- ions as x = 0.0027 M, and solve for the concentration of H+ ions using the equation:
[H+] * [OH-] = 10^(-14)
[H+] = 10^(-14) / 0.0027
[H+] ≈ 3.7 * 10^(-12) M
Finally, we can calculate the pH using the concentration of H+ ions:
pH = -log10 (3.7 * 10^(-12))
pH ≈ 11.4
Therefore, the pH of a sodium ethanoate solution with a concentration of 0.1 M is approximately 11.4.