What is the pH of a 0.15M HY solution with a pKa value of 3 and Ka value of 1e-3?

**Answer 1**

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