how would you do this question
when i did it i did -log(1.0) but that is just 0 so what it the procedure to do this question
Calculate the pH of a 1.0 mol/L aqueous solution of sodium benzoate. Note: Only the benzoate ion affects the pH of the solution.
Benzoate ion hydrolyzes in water to form a solution that is not neutral.
If we call the benzoate ion Bz^-, then
.........Bz^- + HOH ==> HBz + OH^-
I........1.0..............0.....0
C........-x..............x.....x
E.......1-x..............x......x
Kb for Bz^- = (Kw/Ka for HBz) = x*x/(1-x) and solve for x = (OH^-),then convert to pH.
HBz is benzoic acid. You can find Ka for HBz in you text/notes I'm sure but it's approx 10^-5.
thanks
To calculate the pH of a solution of sodium benzoate, you need to consider the dissociation of sodium benzoate in water.
The equation for the dissociation of sodium benzoate (C6H5COONa) is as follows:
C6H5COONa(s) → C6H5COO-(aq) + Na+(aq)
As per the given question, only the benzoate ion (C6H5COO-) affects the pH. This is because Na+(aq) is the spectator ion and does not participate in the acid-base reaction. Hence, you need to consider the hydrolysis reaction of the benzoate ion.
The hydrolysis reaction of the benzoate ion (C6H5COO-) with water (H2O) is as follows:
C6H5COO-(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH-(aq)
From the hydrolysis reaction, you can see that benzoate ions react with water to produce benzoic acid (C6H5COOH) and hydroxide ions (OH-).
Now, to find the pH of the solution:
Step 1: Calculate the concentration of OH- ions
Since the solution is 1.0 mol/L, the concentration of the benzoate ions is also 1.0 mol/L.
However, for every benzoate ion that hydrolyzes, an equal amount of hydroxide ion is produced. Therefore, the concentration of OH- ions is the same as the concentration of the benzoate ions, which is 1.0 mol/L.
Step 2: Calculate the pOH
pOH = -log[OH-]
pOH = -log(1.0) = 0
Step 3: Calculate the pH
pH + pOH = 14 (at room temperature)
So, pH = 14 - pOH
pH = 14 - 0 = 14
Therefore, the pH of the 1.0 mol/L aqueous solution of sodium benzoate is 14.