Sunday

December 21, 2014

December 21, 2014

Posted by **Jake** on Wednesday, May 16, 2012 at 2:53pm.

Ka = 6.3 x 10^-5

- Chemistry -
**DrBob222**, Wednesday, May 16, 2012 at 3:07pmmols 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.

- Chemistry -
**Jake**, Wednesday, May 16, 2012 at 3:13pmMols 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..

- Chemistry -
**DrBob222**, Wednesday, May 16, 2012 at 3:44pmI 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.

- Chemistry -
**Jake**, Wednesday, May 16, 2012 at 3:53pmThanks so much, you're an awesome help.

**Answer this Question**

**Related Questions**

chemistry - What is the pH of a solution prepared by dissolving 12.2 g of ...

Chemistry (reposts#2) - A solution is prepared by dissolving 5.00 g of ferric ...

Chemistry - 1. Calculate the pH of a buffer solution that contains 0.32 M ...

Chemistry - 1. Calculate the pH of a buffer solution that contains 0.32 M ...

Chemistry - Calculate the molarity of Na2C204 solution prepared by dissolving 1....

Chemistry - Calculate the molarity of Na2C204 solution prepared by dissolving 1....

chemistry - A solution is prepared by dissolving 0.5842g of oxalic acid (H2C2O4...

chemistry - A standard solution is prepared for the analysis of fluoxymesterone...

Chemistry - You have 150 ml of a solution of benzoic acid in water estimated to ...

Chemistry - You have 150 ml of a solution of benzoic acid in water estimated to ...