An organic acid concentration in a hard sample was determined by diluting a 15.00 mL sample to 200.00 mL solution. A 20.00 mL portion of the diluted sample reacted with 40.00 mL of 0.05417 M NaOH to completion. The excess OH- was back-titrated with 5.21 mL of 0.02154 M HCl. The concentration of the organic acid in the original sample is:
a) 0.68 M
b) 1.37 M
c) 0.137 M
d) 13.7 M
A friend showed me this problem, and I know the answer is [ b) 1.37 M ], but I can't seem to figure out the process behind it.
I'm able to determine 2.267 mmol of NaOH is titrated, and 0.1122 mmol of HCl is back-titrated, meaning (I would guess) 2.055 mmol Organic Acid was in the diluted sample.
Multiply this by 10 to get from 20 mL back to 200 mL, then multiply by 40/3 to counteract the dilution from 15.00 mL to 200.00 mL.
My answer: .274 M Organic Acid
I realize I've gone wrong somewhere, but have hit a wall -- can anyone point me in the correct direction? Thank you for any help!
a) 0.68 M
b) 1.37 M
c) 0.137 M
d) 13.7 M
A friend showed me this problem, and I know the answer is [ b) 1.37 M ], but I can't seem to figure out the process behind it.
I'm able to determine 2.267 mmol of NaOH is titrated, and 0.1122 mmol of HCl is back-titrated, meaning (I would guess) 2.055 mmol Organic Acid was in the diluted sample.
Multiply this by 10 to get from 20 mL back to 200 mL, then multiply by 40/3 to counteract the dilution from 15.00 mL to 200.00 mL.
My answer: .274 M Organic Acid
I realize I've gone wrong somewhere, but have hit a wall -- can anyone point me in the correct direction? Thank you for any help!
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Work backwards and reason this out. The final step in the problem neutralizes the excess OH.
(5.21mL / 1000 mL per L) (.02154 M HCl) = 1.122234 x 10 -4 mol HCl <--- this is the moles of acid required to neutralize the the base, we are lucky because strong acids and bases are used and dissociate completely and thus simplify the problem.
1.122234 x 10 -4 mol HCl = 1.122234 x 10 -4 mol NaOH that was in excess from reaction with the organic acid.
The total amount of NaOH to used react with organic acid was:
.04 x .05417 M NaOH = .0021668 mol NaOH
Since we have the excess amount at the end and original amount used, we can find the amount of NaOH that reacted with the organic acid (have to assume the organic acid dissociates completely).
.0021668 mol NaOH - 1.122234 x 10 -4 mol NaOH = .0020545766 mol NaOH that reacted with diluted organic acid.
The amount of moles of organic acid that reacted with NaOH = .0020545766 (mol Organic Acid)
.0020545766 mol Organic Acid in 20 mL = .10272883 M
This is the concentration of the 20 mL portion and diluted sample. From that you calculate .020545766 mol of Organic Acid in 200 mL. Take that mole amount and now divide by .015 L to get your answer.
.020545766 mol Organic Acid / .015 L = 1.3697177 M Organic Acid
1.37 M Organic Acid
(5.21mL / 1000 mL per L) (.02154 M HCl) = 1.122234 x 10 -4 mol HCl <--- this is the moles of acid required to neutralize the the base, we are lucky because strong acids and bases are used and dissociate completely and thus simplify the problem.
1.122234 x 10 -4 mol HCl = 1.122234 x 10 -4 mol NaOH that was in excess from reaction with the organic acid.
The total amount of NaOH to used react with organic acid was:
.04 x .05417 M NaOH = .0021668 mol NaOH
Since we have the excess amount at the end and original amount used, we can find the amount of NaOH that reacted with the organic acid (have to assume the organic acid dissociates completely).
.0021668 mol NaOH - 1.122234 x 10 -4 mol NaOH = .0020545766 mol NaOH that reacted with diluted organic acid.
The amount of moles of organic acid that reacted with NaOH = .0020545766 (mol Organic Acid)
.0020545766 mol Organic Acid in 20 mL = .10272883 M
This is the concentration of the 20 mL portion and diluted sample. From that you calculate .020545766 mol of Organic Acid in 200 mL. Take that mole amount and now divide by .015 L to get your answer.
.020545766 mol Organic Acid / .015 L = 1.3697177 M Organic Acid
1.37 M Organic Acid
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No problem. Glad I could help. Good luck with your work.
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