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What is the pH of the buffer that results when 0.235 mol of NH3 and (2.46x10^-1)...

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Chemistry

What is the pH of the buffer that results when 0.235 mol of NH3and (2.46x10^-1) mol of NH4Cl are dissolved in water to a volume of0.50 L? (Ka of NH4(+) = 5.6E-10)

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4.1 Ratings (500 Votes)

Ka of NH4+ = 5.6 x 10^-10

Ka x Kb = Kw = 1.0 x 10^-14

Kb of NH4+ = 1.0 x 10^-14 / 5.6 x 10^-10

                    = 1.78 x 10^-5

pKb = -log Kb

       = -log (1.78 x 10^-5)

      = 4.75

NH3 molarity (base)= 0.235 / 0.5

                  = 0.47 M

NH4Cl molarity (salt)= 2.46x10^-1 / 0.5

                      = 0.492 M

pOH = pKb + log (salt / base)

        = 4.75 + log [0.492 / 0.47]

        = 4.77

pH + pOH = 14

pH = 14 - pOH

     = 14 - 4.77

    = 9.23

pH of the buffer = 9.23
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