. (a) If your titration solution is 0.444 M in NaOH, and the endpoint occurs at

Question

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(a) If your titration solution is 0.444 M in NaOH, and the endpoint occurs at 13.40 mL of titrant, how many mmol of NaOH are required to reach the endpoint?
b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH?
(c) How many grams of acetic acid is this? WebAssign will check your answer for the correct number of significant figures. grams HC2H3O2
(d) If the mass of analyte is 10.30 grams, what is the mass % of acetic acid in the analyte?
.
(a) If your titration solution is 0.444 M in NaOH, and the endpoint occurs at 13.40 mL of titrant, how many mmol of NaOH are required to reach the endpoint?
b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH?
(c) How many grams of acetic acid is this? WebAssign will check your answer for the correct number of significant figures. grams HC2H3O2
(d) If the mass of analyte is 10.30 grams, what is the mass % of acetic acid in the analyte?
.
(a) If your titration solution is 0.444 M in NaOH, and the endpoint occurs at 13.40 mL of titrant, how many mmol of NaOH are required to reach the endpoint?
b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH?
(c) How many grams of acetic acid is this? WebAssign will check your answer for the correct number of significant figures. grams HC2H3O2
(d) If the mass of analyte is 10.30 grams, what is the mass % of acetic acid in the analyte?

Explanation / Answer

a) No of milli mol = 0.444*13.4 = 5.95 milimol

b) No of milli mol of CH3COOH = 5.95 milimol

C) mass of CH3COOH = 5.95*10^-3*60 = 0.357 grams

d) % of CH3COOH = 0.357/10.3*100 = 3.47%

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