The pKa for (NH4+) is 9.24. You have 600 mL of each of the following four buffer

Question

The pKa for (NH4+) is 9.24. You have 600 mL of each of the following four buffers available:

i) 0.1M NH3 and 0.2 M NH4Cl

ii) 0.025M NH3 and 0.05 M NH4Cl

iii) 0.025M NH3 and 0.025 M NH4Cl

iv) 0.05M NH3 and 0.05 M NH4Cl

a) Calculate the pH of each buffer.

b) What would the pH of each buffer be after adding either 200 mL, 400 mL, or 800 mL of a 0.01M HCl (the volume of HCl is not cumulative, i.e., calculate the pH for adding 200 mL of 0.01 M HCl to 600 mL of the first buffer, then calculate the pH for adding 400 mL of 0.01M HCl to 600mL of the first buffer etc; do this for all buffers.)

c) What volume of the 0.01 M HCl does need to be added to each buffer to change the pH by one unit.

d) Which buffer has the higher buffest capacity and which one the lowest? Ka(NH4+)=5.70·10^-10

Explanation / Answer

i) 0.1M NH3 and 0.2 M NH4Cl

pKa + pKb = 14

pKb = 4.76

for basic buffer

pH = 14 - {pKb+ log[NH4Cl]/[NH3]}

pH =14 - (4.76 + log (0.2 /0.1))

pH = 8.94

(ii) 0.025M NH3 and 0.05 M NH4Cl

pH =14 - (4.76 + log (0.05 / 0.025))

pH = 8.94

(iii) 0.025M NH3 and 0.025 M NH4Cl

pH = 14 - (4.76 + log (0.025 / 0.025))

pH = 9.24

(iv) 0.05M NH3 and 0.05 M NH4Cl

pH = 14 - (4.76 + log (0.05 / 0.05))

pH = 9.24

b) 0.1M NH3 and 0.2 M NH4Cl

200ml 0.01M HCl added to this buffer .

millimoles of HCl = C = 200 x 0.01 = 2

millimoles of   NH4Cl = 0.2 x 600 = 120

millimoles of NH3 = 0.1 x 600 = 60

on addition of acid HCl to the basic buffer salt moles incrases and base moles decreases

pH = 14 - {pKb+ log[ NH4Cl + C ]   / [ NH3 -C] }

pH =14 - (4.76 + log (120-2 / 60 +2))

pH = 8.98

all problems are same way try to solve please. i could maximum do these only according to our chegg rules

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