Question
A 0.25 mol/L solution of benzoic acid, KC7H5O2 and antiseptic also used as a food preservative, has a pH of 2.40. Calculate the Ka of benzoic acid at SATP
My work:
C6H5COOH --> C6H5COO^- + H^+
Ka = (C6H5COO-)(H+)/ C6H5COOH
I know my next steps are to use pH = log(H+) to calculate H+
and then
substitute into Ka expression and determine Ka
How to do this?
Thanks in advance
My work:
C6H5COOH --> C6H5COO^- + H^+
Ka = (C6H5COO-)(H+)/ C6H5COOH
I know my next steps are to use pH = log(H+) to calculate H+
and then
substitute into Ka expression and determine Ka
How to do this?
Thanks in advance
Answers
DrBob222
pH = 2.40
pH = -log(H+)
-log(H^+) = 2.40
log(H^+) = -2.40
plug -2.40 into your calculator and hit the 10<sup>x</sup> key to obtain 0.00398. check my work.
pH = -log(H+)
-log(H^+) = 2.40
log(H^+) = -2.40
plug -2.40 into your calculator and hit the 10<sup>x</sup> key to obtain 0.00398. check my work.
Ross
0.003898-- this is H+?
Ka = ( )(0.00398)/( )
What would be the concentration of C6H5COO- ?
Ka = ( )(0.00398)/( )
What would be the concentration of C6H5COO- ?
DrBob222
Look at the equation.
C6H5COOH ==> C6H5COO^- + H^+
For every <b>1 mol C6H5COOH</b> that ionizes, you get <b>1 mol C6H5COO^-</b> and <b> 1 mol H^+</b>. So they are the same.
C6H5COOH ==> C6H5COO^- + H^+
For every <b>1 mol C6H5COOH</b> that ionizes, you get <b>1 mol C6H5COO^-</b> and <b> 1 mol H^+</b>. So they are the same.
Kelsy
I get 4.0 x 10^-3 but im incorrect
DrBob222
No.
(0.00398)^2/(0.25-0.00398) =
6.44 x 10^-5
(0.00398)^2/(0.25-0.00398) =
6.44 x 10^-5