What is the pH of a 0.15M HY solution with a pKa value of 3 and Ka value of 1e-3?
-
HY <----> H+ + Y-
initially
0.15 <-----> 0 + 0
at equilibrium...
0.15-x <------> x + x
Ka = [H+] [Y-] / [Ha]
putting the values...
10^-3 = x X x / (0.15-x)
10^-3 = x^2 / (0.15-x)
x^2 = 10^-3(0.15-x)
x^2 = 1.5X10^-4 - 10^-3x
x^2 + 10^-3x - 1.5 X 10^-4 = 0
this is a quadratic equation ....
x = [ -b +_ square root of (b^2-4ac) ] / 2a
as x here represents concentration so it cant be negative so there will be only one solution ..
x = [-b + square root of (b^2-4ac) ] / 2
where a = 1
b = 10^-3
c = -1.5 X 10^-4
x = [ -10^-3 +_ square root of ( (10^-3)^2 - 4 X 1 X -1.5 X 10^-4) ] / 2
x = [ -10^-3 + square root of ( 10^-6 + 6 X 10^-4 ) ] / 2
x = [-10^-3 + square root of 6 X 10^-4 ]/2
x = [ -10^-3 + 0.0244 ] /2
x = 0.0234/2 = 0.0117 M
so conc. of H+ = 0.0117 M
pH = -log [H+] = -log 0.0117 = 1.932
please check the maths and feel free to ask any questions
initially
0.15 <-----> 0 + 0
at equilibrium...
0.15-x <------> x + x
Ka = [H+] [Y-] / [Ha]
putting the values...
10^-3 = x X x / (0.15-x)
10^-3 = x^2 / (0.15-x)
x^2 = 10^-3(0.15-x)
x^2 = 1.5X10^-4 - 10^-3x
x^2 + 10^-3x - 1.5 X 10^-4 = 0
this is a quadratic equation ....
x = [ -b +_ square root of (b^2-4ac) ] / 2a
as x here represents concentration so it cant be negative so there will be only one solution ..
x = [-b + square root of (b^2-4ac) ] / 2
where a = 1
b = 10^-3
c = -1.5 X 10^-4
x = [ -10^-3 +_ square root of ( (10^-3)^2 - 4 X 1 X -1.5 X 10^-4) ] / 2
x = [ -10^-3 + square root of ( 10^-6 + 6 X 10^-4 ) ] / 2
x = [-10^-3 + square root of 6 X 10^-4 ]/2
x = [ -10^-3 + 0.0244 ] /2
x = 0.0234/2 = 0.0117 M
so conc. of H+ = 0.0117 M
pH = -log [H+] = -log 0.0117 = 1.932
please check the maths and feel free to ask any questions