Question

# If a solution is 0.10 M phosphoric acid, what is the solution pH? Ka1 ​ =...

If a solution is 0.10 M phosphoric acid, what is the solution pH? Ka1 ​ = 0.0075.

H3PO4 dissociates as:

H3PO4 -----> H+ + H2PO4-

0.1 0 0

0.1-x x x

Ka = [H+][H2PO4-]/[H3PO4]

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 ((7.5*10^-3)*0.1) = 2.739*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

7.5*10^-3 = x^2/(0.1-x)

7.5*10^-4 - 7.5*10^-3 *x = x^2

x^2 + 7.5*10^-3 *x-7.5*10^-4 = 0

a = 1

b = 7.5*10^-3

c = -7.5*10^-4

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 = 3.056*10^-3

roots are :

x = 2.389*10^-2 and x = -3.139*10^-2

since x can't be negative, the possible value of x is

x = 2.389*10^-2

So, [H+] = x = 2.389*10^-2 M

use:

pH = -log [H+]

= -log (2.389*10^-2)

= 1.6218

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