Question

Calculate the change in pH when 0.334 mol H+ is added to 1.00 L of each of the following buffers.

a 0.580 M solution of pyridine (py) containing 0.500 M pyH+

a 0.580 M solution of aniline (an) containing 0.900 M anH+

Answer 1

pH of buffer is given by Henderson’s equation as:

pH = pKa + log{[base]/[conjugate acid]}

if pH₁ and pH₂ be two different pH at different Base and conjugate acid conditions then,

pH₁ = pKa + log{[base]₁ / [conjugate acid]₁}

pH₂ = pKa + log{[base]₂ / [conjugate acid]₂}

Then change in pH will be given as,

pH₁ - pH₂ = log{[base]₁ / [conjugate acid]₁} - log{[base]₂ / [conjugate acid]₂} ………… (1)

1) For Pyridine/Pyridinium buffer,

Initially, [Py]1 = 0.58 M, [PyH⁺]1 = 0.50 M

Number of moles of HCl added = 0.334 moles/L [ As Buffer volume is 1L]

Addition of H⁺ ions decrease concentration of base [Py] by 0.334 M and increase that of [PyH⁺] by 0.334 M

Hence finally, [Py]2 = 0.58 - 0.334 = 0.246 M

[PyH⁺]2 = 0.50 + 0.334 = 0.834 M

Eq. (1) for Pyridine/Pyridinium buffer can be written as,

pH1 - pH2 = log{[Py]1 /[ PyH⁺]1} - log{[Py]2 /[ PyH⁺]2}

pH1 - pH2 = log(0.58/0.50) – log(0.246/0.834)

pH1 - pH2 = 0.0644 - (-0.531)

pH1 – pH2 = 0.596

Change in pH is 0.596

i.e. after addition of 0.334 mol H⁺ pH will decrease by 0.596 unit.

2) For Aniline/Anilinium buffer,

Initially, [An]1 = 0.58 M, [AnH+]1 = 0.90 M

Number of moles of HCl added = 0.334 moles/L [ As Buffer volume is 1L]

Addition of H⁺ ions decrease concentration of base [An] by 0.334 M and increase that of [AnH+] by 0.334 M

Hence finally, [An]2 = 0.58 -0.334 = 0.246 M

[PyH⁺]2 = 0.90 + 0.334 = 1.234 M.

Eq. (1) for Pyridine/Pyridinium buffer can be written as,

pH1 - pH2 = log{[An]1 / [ AnH+]1} - log{[An]2 / [ AnH+]2}

pH1 - pH2 = log(0.58/0.90) – log(0.246/1.234)

pH1 - pH2 = (-0.19) – (-0.70)

pH1 – pH2 = 0.51

Change in pH is 0.51

i.e. after addition of 0.334 mol H⁺ pH will decrease by 0.51 unit.