Question

Calculate the pH of the solution that results from each of the following mixtures:

A) 55.0 mL of 0.16 M HCHO2 with 75.0mL of 0.13M NaCHO2

B) 130.0mL of 0.11 M NH3 with 260.0 mL of 0.11 M NH4Cl

Answer 1

pH of acidic buffer and pOH of basic buffer is given by the Henderson-Hasselbalch equation,

pH = pKa + log ([Conjugate base]/[Acid]) ........... (for acidic buffer)

pOH = pKb + log([Conjugate acid]/[Base]) ............. (For basic buffer)

Here,

pKa and pKb are for acid and base respectively and concentrations like [Conjugate base] , [Acid] are in M/L unit for separate solutions i.e. before mixing to form buffer.

We will not conver concentrations like [Conjugate base], [Acid] etc in M/L unit which involve finding milimoles and division by 1000 mL. 1000 factor going to come for both and will be mutually cancelled out. So we will find # of milimoles for each species involved and will use it to find ration like [Conjugate base]/[Acid] and proccee with calculation.

A) For Formic acid (HCHO2) - Sodium formate (NaCHO2) buffer we have,

55.0 mL of 0.16 M HCHO2

# of milimoles of formic acid = 0.16 x 55.0 = 8.80

and 75.0mL of 0.13M NaCHO2

# of milimoles of sodium formate = 0.13 x 75.0 = 9.75

pKa of formic acid = 3.75

Then using Henderson-Hasselbalch equation,

pH = pKa + log([Sodium formate]/[Formic acid])

pH = 3.75 + log(9.75 / 8.80)

pH = 3.75 + 0.05

pH = 3.80

pH of buffer formed using 55.0 mL of 0.16 M HCHO2 with 75.0mL of 0.13M NaCHO2 is 3.80.

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B) For a basic buffer Ammonia NH3 - Ammonium chloride NH4Cl.

130.0mL of 0.11 M NH3

# of milimoles of NH3 = 0.11 x 130.0 = 14.3

and 260.0 mL of 0.11 M NH4Cl

# of milimoles of NH4Cl = 0.11 x 260 = 28.6

pKb of NH3 = 4.75

Then using Henderson-Hasselbalch equation for pOH of basic buffer we write,

pOH = pKb + log([NH4Cl]/[NH3])

pOH = 4.75 + log(28.6/14.3)

pOH = 4.75 + 0.30

pOH = 5.05

Then, pH + pOH = 14

pH + 5.05 = 14

pH = 14 - 5.05

pH = 8.95

pH of buffer formed from 130.0mL of 0.11 M NH3 and 260.0 mL of 0.11 M NH4Cl is 8.95.

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