Calculate the pH of a solution prepared by dissolving 12.2 g of benzoic acid in enough water to produce a 500 mL solution.

Ka = 6.3 x 10^-5

I assume up to this you understand.

(x)(x)/0.1998-x = 6.3E-5.
First let's round those numbers to make it less confusing.
6.3E-5 = x^2/0.2-x.
We make the assumption that 0.2-x = 0.2 (that is that x is so small that subtracted from 0.2 makes almost no difference. We can check the assumption later.)
6.3E-5 = x^2/0.2
6.3E-5*0.2 = x^2
1.26E-5 = x^2
x = sqrt(1.26E-5) = 3.55E-3 = (H^+)
pH = -log(H^+) = -log(3.55E-3) = -(-2.45) = 2.45
Three important things here.
#1. You could have not made the assumption that 0.2-x = 0.2 which would have led to a quadratic equation. You could have solved that but making the assumption avoids the quadratic and the answer is essentially the same.
#2. We can check to see if the assumption of 0.2-x = 0.2 is ok. How do we do that? Our provisional answer is 3.55E-3 so 0.2-0.00345 for all practical purposes = 0.2 and we make no mistake by making the assumption BUT it saves us the trouble of solving a quadratic. Had the check shown that the assumption was not valid we would have had to go back and solve the quadratic equation.
#3. Your "corrections" above are not valid although they are correct. 12.2 g has 3 significant figures so 0.0999 is ok. Then 0.0999/0.5 = 0.1998M but you are allowed only 3 places so that rounds to 0.200 M.
I hope this helps your confusion.

mols HBz = 12.2g/molar mass = about 0.1

M HBz = 0.1/0.5L = 0.2M
............HBz ==> H^+ + Bz^-
I...........0.2......0.....0
C..........-x........x......x
E........0.2-x.......x......x
Substitute the E line into the Ka expression and solve for x, then convert to pH.

Thanks so much, you're an awesome help.

To calculate the pH of the solution, we first need to determine the concentration of benzoic acid in the solution. We can do this by using the given mass of benzoic acid and the volume of the solution.

1. Calculate the moles of benzoic acid:
- To do this, we need to know the molar mass of benzoic acid. The chemical formula for benzoic acid is C6H5COOH, and its molar mass is 122.12 g/mol.
- Divide the given mass of benzoic acid (12.2 g) by the molar mass to get the number of moles:
Moles = Mass / Molar mass
= 12.2 g / 122.12 g/mol
≈ 0.100 moles

2. Calculate the concentration of benzoic acid in the solution:
- The volume of the solution is given as 500 mL. However, pH calculations usually require concentrations in moles per liter (M).
- Convert the volume from milliliters to liters:
Volume (liters) = 500 mL / 1000
= 0.500 L

- Divide the moles of benzoic acid by the volume in liters to get the concentration in moles per liter:
Concentration (M) = Moles / Volume (liters)
= 0.100 moles / 0.500 L
= 0.200 M

3. Use the concentration to find the concentration of H+ ions:
- Benzoic acid is a weak acid, and its ionization is determined by its acid dissociation constant (Ka). The formula for benzoic acid ionization is:
C6H5COOH ⇌ C6H5COO- + H+

- The equation for Ka is:
Ka = [C6H5COO-][H+] / [C6H5COOH]

- We can assume that since benzoic acid is a weak acid, the concentration of C6H5COO- will be equal to the concentration of H+ ions. Therefore:
Ka = [H+]^2 / [C6H5COOH]
[H+]^2 = Ka x [C6H5COOH]
[H+]^2 = (6.3 x 10^-5) x (0.200)

- Take the square root of both sides to find [H+]:
[H+] = sqrt((6.3 x 10^-5) x (0.200))

4. Calculate the pH:
- The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions.
pH = -log10([H+])

- Substitute the calculated value of [H+] into the equation to get the pH:
pH = -log10(sqrt((6.3 x 10^-5) x (0.200)))

Calculating this equation will give you the pH value.

Mols benzoic acid = 12.2g/122.12g = 0.0999

Int. Concentration of Benzoic acid = 0.0999 mol/ 0.500L = 0.1998 M
Ka = 6.3 x 10-5 = (x)(x)/0.1998 – x
X = [H+] = * M
pH = -log(*) = *

Sorry, this question is just really confusing me..