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Calculate the pH of a buffer solution after the addition of 15.00 mL of 0.100 M...

Calculate the pH of a buffer solution after the addition of 15.00 mL of 0.100 M NaOH to 50.00mL of the buffer that originally contains 0.100 M propanoic acid and 0.0750 M Sodium propionate. The Ka for propanoic acid is 1.3x10-5.

Expert Solution

CONCEPT: CH3CH2COOH + NaOH ---------> CH3CH2COONa + H2O

from above chemical equation it obvious that one mole of propanoic acid reacts with one mole of sodium hydroxide to produce one mole of sodium propanoiatepH of buffer is determined by followin equation.

we have to determine amount of propanoic acid consumed and amount od sodium propanoate formed then calculate their molarity and use in following equation.

pH = pKa + log [Sodium propionate] / [propanoic acid]

pKa = -logKa

pKa = -log 1.3x10^-5

pKa = 4.87

moles of NaOH added = molarity X volume

= 0.1 X0.015 15mL = 0.015L

= 0.0015 Moles

0.0015 Moles of sodium hydroxide will react with 0.0015 moles of propanoic acid and 0.0015 moles of sosium propanoiate will be formed

Mols of propanoic acid origenally present = 0.1 X 0.05 = 0.005 Moles

therefore moles of propanoic acid left = 0.005 - 0.0015 = 0.0035 moles

Molarity of propanoic acid = no. of moles / volume in liters

= 0.0035 / 0.065= 0.054M

Because total volume = 15mL + 50mL = 65mL = 0.065L

Original moles of Sodium propionate = 0.075 X 0.05 = 0.00375 Mols

moles of Sodium propionate after the reaction = 0.00375 + 0.0015 = 0.00525 moles

Molarity = 0.00525 / 0.065 = 0.081M

use the values in boxes in pH equation shown above

pH = 4.87 + log [0.081] / [0.054]

pH = 4.956

queen_honey_blossom answered 1 hour ago

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