Question

2. What is the pH of 3 mM H3PO4 solution? If you wish to adjust the pH of 1 L of that solution to 7.4, how much H+ or OH— do you need (in mmol)? At pH 7.4 what is the concentration of H2PO4 —? The pKas of H3PO4 are 2.1, 7.1 and 12.3.

Answer 1

Solution:

a. What is the pH of 3 mM H3PO4 solution?

             H₃PO₄     +    H₂O          --->         H₂PO₄^-     +    H₃O⁺

i:          3x10^-3 M

c:                -x                                               +x                  +x

e:          3x10^-3 M -x                                +x                  +x

         

         

         rearranging,

   x² + 7.9433x10^-3x -2.3830x10^-5 = 0

use quadratic equation in a calculator and input the values

x = 2.3215x10^-3 M= [H₃O⁺]

   pH = -log [H₃O⁺] = -log (2.3215x10^-3 M) = 2.63

b. If you wish to adjust the pH of 1 L of that solution to 7.4, how much H+ or OH— do you need (in mmol)?

   the initial pH of the solution is 2.63 and contains 2.3215x10^-3 M H₃O⁺, so to get a pH of 7.4 and which contains 3.9811x10^-8 M H₃O⁺ we deduct 3.9811x10^-8 M from 2.3215x10^-3 M,

  H₃O⁺              + OH^-    ---->      H₃O⁺     

         2.3215x10^-3 M    x    3.9811x10^-8 M

         x = [OH^-] = 2.3215x10^-3 M - 3.9811x10^-8 M = 2.32146x10^-3 M or 2.32146 mM

c. At pH 7.4 what is the concentration of H2PO4^-?

        at pH 7.4, [H₃O⁺ ] = [HPO4^2-] = 3.9811x10^-8 M

         pH = pKa + log ([HPO4^2-]/[H2PO4^-])

         7.4 = 7.1 + log ( 3.9811x10^-8 M/[H2PO4^-])

[H2PO4^-] = 3.9811x10^-8 M / 1.995 = 1.996x10^-8 M