# Chemistry

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

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1. 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.

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2. 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..

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3. 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.

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4. Thanks so much, you're an awesome help.

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