analytical chemist is titrating 55.5mL of a 1.100M solution of butanoic acid HC3H7CO2 with a 0.9700M solution of NaOH. The pKa of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 69.8mL of the NaOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added.Round your answer to 2 decimal places.
Question
analytical chemist is titrating 55.5mL of a 1.100M solution of butanoic acid HC3H7CO2 with a 0.9700M solution of NaOH. The pKa of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 69.8mL of the NaOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added.Round your answer to 2 decimal places.
Solution
To solve this problem, we need to follow these steps:
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Calculate the moles of butanoic acid and NaOH.
Moles of butanoic acid = volume (L) x concentration (M) = 0.0555 L x 1.100 M = 0.06105 mol Moles of NaOH = volume (L) x concentration (M) = 0.0698 L x 0.9700 M = 0.067746 mol
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Determine the limiting reactant. In this case, NaOH is in excess and butanoic acid is the limiting reactant.
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Calculate the moles of butanoic acid that remain after the reaction with NaOH.
Moles of butanoic acid remaining = initial moles of butanoic acid - moles of NaOH = 0.06105 mol - 0.067746 mol = -0.006696 mol
Since we can't have negative moles, this means all the butanoic acid has reacted and we have excess NaOH.
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Calculate the moles of NaOH that remain after the reaction.
Moles of NaOH remaining = initial moles of NaOH - moles of butanoic acid = 0.067746 mol - 0.06105 mol = 0.006696 mol
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Calculate the concentration of NaOH in the final solution.
Concentration of NaOH = moles of NaOH / total volume (L) = 0.006696 mol / (0.0555 L + 0.0698 L) = 0.053 M
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Since NaOH is a strong base, it will completely dissociate in water, so the pH of the solution will be determined by the pOH of the NaOH.
pOH = -log[OH-] = -log(0.053) = 1.28
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Convert pOH to pH.
pH = 14 - pOH = 14 - 1.28 = 12.72
So, the pH of the solution after the chemist has added 69.8 mL of the NaOH solution to it is 12.72.
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