Question

SOH You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.075 M N

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Answer #1

fraction as LH+ = 0.646

Explanation

concentration MOPS = 0.100 M

volume MOPS = 10.0 mL

moles MOPS = (concentration MOPS) * (volume MOPS)

moles MOPS = (0.100 M) * (10.0 mL)

moles MOPS = 1.00 mmol

moles NaOH = (concentration NaOH) * (volume NaOH)

moles NaOH = (0.075 M) * (10.0 mL)

moles NaOH = 0.75 mmol

moles conjugate base formed = moles NaOH added

moles conjugate base formed = 0.75 mmol

moles MOPS remaining = (initial moles MOPS) - (moles conjugate base formed)

moles MOPS remaining = (1.00 mmol) - (0.75 mmol)

moles MOPS remaining = 0.25 mmol

Ka MOPS = 6.3 x 10-8

pKa = -log(Ka)

pKa = -log(6.3 x 10-8)

pKa = 7.20

According to Henderson - Hasselbalch equation,

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

pH = pKa + log(moles conjugate base formed / moles MOPS remaining)

pH = 7.20 + log(0.75 mmol / 0.25 mmol)

pH = 7.68

pOH = 14 - pH

pOH = 14 - 7.68

pH = 6.32222

Kb lidocaine = 8.7 x 10-7

pKb = -log(Kb)

pKb = -log(8.7 x 10-7)

pKb = 6.06

According to Henderson - Hasselbalch equation,

pOH = pKb + log([conjugate acid] / [weak base])

pOH = pKb + log([LH+] / [L])

6.32222 = 6.06 + log([LH+] / [L])

log([LH+] / [L]) = 6.32222 - 6.06

log([LH+] / [L]) = 0.26

[LH+] / [L] = 100.26

[LH+] / [L] = 1.827

[L] / [LH+] = 1/1.827

[L] / [LH+] = 0.547

Adding 1 to both sides,

([L] + [LH+]) / [LH+] = 1.547

total concentration / [LH+] = 1.547

taking reciprocal,

[LH+] / total concentration = 1/1.547

[LH+] / total concentration = 0.646

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