what should the pH of 0.1 M sodium acetate be?
Ka = 1.8*10^-5
Answers
Explanation:
First off, you need to recognize that sodium acetate (AcONa) is a base. When dissolved in water, the equilibrium shown below is established.
AcONa = AcOH + OH-
The equilibrium expression is
Kb = [AcOH][OH-]/[AcONa]
Since AcONa is a weak base all three species will be present in solution. Some AcONa will be converted to AcOH and OH-. Let ‘x’ be the amount that is converted, [AcONa] = 0.1 - x, and [AcOH] = [OH-] = x and putting these in to the equilibrium expression gives
Kb = x^2/(0.1-x) => 0.1*Kb -Kb*x = x^2 => x^2 +Kb*x - 0.1*Kb = 0
Now the question gives Ka but you need Kb. Kb = 10^-14/Ka and this translates to Kb = 5.56^-10.
Next you have to use the quadratic formula (x = (-b +/- sqrt(b^2 - 4ac)/2ac) and doing so gives x = 7.46 * 10^-6 and therefore, the concentration of [OH-] = 7.46 * 10^-6.
pOH = - log [OH-] and pH = 14 - p[OH-] so
pH = 14 - (-log[OH-]) => pH = 14 - (-log(7.46 * 10^-6)
pH = 14 - 5.13 => 8.87
The pH I calculate may be a little different from what others calculate due to various rounding errors.
please do mark it as brainliest
