For each of the following solutions, calculate the initial pH and the final pH a

Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH. For 300.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH. Express your answers using three significant figures separated by a comma. For 300.0 mL of a buffer solution that is 0.195 M in HCHO_2 and 0.290 M in KCHO_2, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(K_a = 1.8 10^-4). Express your answers using three significant figures separated by a comma. For 300.0 mL of a buffer solution that is 0.3028 Mm CH_3 CH_2 NH_2 and 0.2783 M in CH_3 CH_2 NH_3 Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(K_b = 6.46 10^-4). Express your answers using three significant figures separated by a comma.

Explanation / Answer

Part A.

Initial pH

pH of water = -log(1 x 10^-7) = 7

After addition of NaOH

molarity of NaOH = moles/L = 0.02/0.3 = 0.067 M

pOH = -log[OH-] = -log(0.067) = 1.174

pH = 14 - pOH = 12.83

Answer : 7 , 12.83

Part B. Initial pH

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

pKa = -logKa = -log(1.8 x 10^-4) = 3.745

Feed values,

pH = 3.745 + log(0.290/0.195)

      = 3.92

Final pH after addition of 0.02 mol of NaOH

initial moles of HCOOH = molarity x volume = 0.195 x 0.3 = 0.0585 mols

initial moles of HCOOK = 0.290 x 0.3 = 0.087 mols

Final moles of HCOOH = 0.0585 - 0.02 = 0.0385 mols

final moles of HCOOK = 0.087 + 0.02 = 0.107 mols

molarity of HCOOH = 0.0385/0.3 = 0.13 M

molarity of HCOOK = 0.107/0.3 = 0.36 M

Feed values in pH equation,

pH = 3.745 + log(0.36/0.13)

     = 4.19

Answer : 3.92 , 4.19

Part C. Initial pH

pKb = -logKb = 3.19

pKa = 14 - pKb = 10.81

pH = 10.81 + log(0.3028/0.2783)

      = 10.85

Final pH

initial moles of CH3CH2NH2 = molarity x volume = 0.3028 x 0.3 = 0.091 mols

initial moles of CH3CH2NH3Cl = 0.2783 x 0.3 = 0.0835 mols

Final moles of CH3CH2NH2 = 0.091 + 0.02 = 0.111 mols

final moles of CH3CH2NH3Cl = 0.0835 - 0.02 = 0.0635 mols

molarity of CH3CH2NH2 = 0.111/0.3 = 0.37 M

molarity of CH3CH2NH3Cl = 0.0635/0.3 = 0.212 M

Feed values in pH equation,

pH = 10.81 + log(0.37/0.212)

     = 11.05

Answer : 10.85 , 11.05

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