C6H5COOH dissociates as:
C6H5COOH -----> H+ + C6H5COO-
0.1 0 0
0.1-x x x
Ka = [H+][C6H5COO-]/[C6H5COOH]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((6.46*10^-5)*0.1) = 2.542*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
6.46*10^-5 = x^2/(0.1-x)
6.46*10^-6 - 6.46*10^-5 *x = x^2
x^2 + 6.46*10^-5 *x-6.46*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.46*10^-5
c = -6.46*10^-6
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 2.584*10^-5
roots are :
x = 2.51*10^-3 and x = -2.574*10^-3
since x can't be negative, the possible value of x is
x = 2.51*10^-3
So, [H+] = x = 2.51*10^-3 M
use:
pH = -log [H+]
= -log (2.51*10^-3)
= 2.6004
Answer: 2.60
IVIUVIT 8 Lu dhouer question will save this response. Question 4 What is the pH of...
What is the pH of a 0.1M solution of benzoic acid? ( Write answer to the hundredths place) KA(benzoic acid) = 6.46x10-5
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A solution of 0.05M hexanoic acid is prepared. KA = 1.41x10-5 a. What is the pH of the solution? (write answer to the hundredths place) b. What is the % ionization of the solution? (write answer to the hundredths place)