Question

If a buffer solution is 0.400 M in a weak acid (Ka = 6.7 × 10-6) and 0.290 M in its conjugate base, what is the pH?

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What is the pH of 75.0 mL of a solution that is 0.041 M in weak base and 0.059 M in the conjugate weak acid (Ka = 7.2 × 10–8)?

Answer 1

P^H    = P^Ka + log[conjugate base]/[acid]

concentration of acid = 0.4M

concentration of conjugate base = 0.29M

P^Ka = -logKa

    = -log(6.7*10^-6)

       = -log6.7+6log10

      = -0.8260+6

P^ka = 5.174

P^H    = 5.174+log0.29/0.4

        = 5.174-0.1396

P^H    = 5.0344 >>>>>>answer

2.    Ka*kb = 10^-14

   Kb      = 10^-14/7.2*10^-8

                Kb = 1.38*10^-7

                            P^Kb   = -log(1.38*10^-7)

                            = -log1.38+7log10

                          = -0.1398+7

                          = 6.8602

                         P^OH = P^Kb + log[conjugate acid]/[base]

             concentration of conjugate acid = 0.059/75

                concentration of base             = 0.041/75

                  p^OH    = 6.8602+log{0.059/75/0.041/75}

                          = 6.8602 + 0.1580

                  P^OH   = 7.0182

                   P^H = 14-P^OH

                          = 14-7.0182

                    P^H = 6.9818 >>>>>answer