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Is buffer capacity constant?

What is the "buffer capacity" of a solution containing .3M acetic acid and its salt?

  • Answer:

    are the acid AND NaAc both 0.3M if so, pKa = pH = 4.76, Ka = 1.74x10^-5, [H+] = 1.74x10^-5M C = (0.3 + 0.3) / 1L = 0.6M B = e x C x Ka x [H+] / (Ka + [H+])^2 B = 2.3 x 0.6M x 1.74x10^-5 x 1.74x10^-5 / (1.74x10^-5 + 1.74x10^-5)^2 B = 0.341

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pH = pKa + log[A-]/HA] If both the acid and salt are 0.3 M, then the pH = pKa which is 4.79. if a small amount of δ of OH- is added, then the pH will be ∆pH = log[(0.3 + δ)/(0.3 - δ)] Let δ = 0.01, then log[(0.3 + δ)/(0.3 + δ)] = 4.81 buffer capacity is ∆[OH-]/∆pH = 0.01/0.029 = 0.34 mol per unit of pH

are the acid AND NaAc both 0.3M if so, pKa = pH = 4.76, Ka = 1.74x10^-5, [H+] = 1.74x10^-5M C = (0.3 + 0.3) / 1L = 0.6M B = e x C x Ka x [H+] / (Ka + [H+])^2 B = 2.3 x 0.6M x 1.74x10^-5 x 1.74x10^-5 / (1.74x10^-5 + 1.74x10^-5)^2 B = 0.341

Caroline Miller

pH = pKa + log[A-]/HA] If both the acid and salt are 0.3 M, then the pH = pKa which is 4.79. if a small amount of δ of OH- is added, then the pH will be ∆pH = log[(0.3 + δ)/(0.3 - δ)] Let δ = 0.01, then log[(0.3 + δ)/(0.3 + δ)] = 4.81 buffer capacity is ∆[OH-]/∆pH = 0.01/0.029 = 0.34 mol per unit of pH

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