Asked by Bella
A solution is prepared by dissolving 50.0 g of pure HC2H3O2 and 20.0 g of NaC2H3O2 in 975 mL of solution (the final volume).
what is the ph?
What would the pH of the solution be if 50.0 mL of 0.900 M NaOH were added?
What would the pH be if 30.0 mL of 0.30 M HCl were added to the original 975 mL of buffer solution?
Please help I have tried using the Henderson Hasselbach equation and I am not getting the correct answer.
what is the ph?
What would the pH of the solution be if 50.0 mL of 0.900 M NaOH were added?
What would the pH be if 30.0 mL of 0.30 M HCl were added to the original 975 mL of buffer solution?
Please help I have tried using the Henderson Hasselbach equation and I am not getting the correct answer.
Answers
Answered by
DrBob222
Show your work and I'll guide you through it.
First, just the pH of the solution.
First, just the pH of the solution.
Answered by
Bella
I multiplied 50g by 975 mL and then 20g by 975 mL. Then I used the given pKa of 4.745 + log (19,500 / 48,750) = 4.35
Answered by
DrBob222
Thos are molarities that go in the HH equation.
mols HAc = 50/molar mass HAc
M HAc = mols/0.975L = ?
mols NaAc = 20/molar mass NaAc.
M NaAc = mols/0.975
pH = pKa + log b/a
I get close to 4.2 but that approximate.
By the way you can take a shortcut and use mols base/mols acid and obtain the same answer as if you used M base/M acid. That's because the 975 mL cancels.
mols HAc = 50/molar mass HAc
M HAc = mols/0.975L = ?
mols NaAc = 20/molar mass NaAc.
M NaAc = mols/0.975
pH = pKa + log b/a
I get close to 4.2 but that approximate.
By the way you can take a shortcut and use mols base/mols acid and obtain the same answer as if you used M base/M acid. That's because the 975 mL cancels.
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