Question

if 10.57 mL of an acetic Acid solution required 10.54 mL 0.9607 M NaOH to reach...

if 10.57 mL of an acetic Acid solution required 10.54 mL 0.9607 M NaOH to reach the endpoint, whats the concentration?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Write the reaction between NaOH and acetic acid

CH3COOH + NaOH --- > CH3COONa + H2O

Mol ratio between NaOH and CH3COOH is 1 : 1

Step 1 calculate moles of NaOH

n NaOH = volume in L x molarity

0.01054*0.9607=0.01013 mol

Step 2 Calculate the moles of CH3COOH

n CH3COOH = moles of NaOH x 1 mol CH3COOH/ 1 mol NaOH

0.01013*1/1=0.01013 mol CH3COOH

Step 3 use volume of acid to calculate concentration

[CH3COOH]= 0.01013 mol / 0.01057 L

0.01013/0.01057=0.96 M

Concentration of acetic acid is 0.96 M

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
if 9.76 mL of an acetic Acid solution required 9.80 mL of 1.0962 M NaOH to...
if 9.76 mL of an acetic Acid solution required 9.80 mL of 1.0962 M NaOH to reach the endpoint, whats the concentration of the Acetic Avid Solution?
A NaOH solution is standardized using KHP and the molarity of the NaOH solution is determined...
A NaOH solution is standardized using KHP and the molarity of the NaOH solution is determined to be 0.4150 M. Since the acid in vinegar is the monoprotic acid, acetic acid, the concentration of the acetic acid can easily be determined by titration. If 90.72 mL of this solution is required to titrate 10.32 mL of vinegar to the Phenolphthalein endpoint, what is the concentration of acetic acid in the vinegar? _______________ M 95.71 mL of NaOH solution is required...
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH....
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH. To reach the endpoint of the titration, 30.00 mL of NaOH solution is required. Ka = 1.8 x 10-4 What is the concentration of formic acid in the original solution? What is the pH of the formic acid solution before the titration begins (before the addition of any NaOH)?
Concentraion of NaOH = 0.1023 M Volume of acetic acid = 20 ml Volume of NaOH...
Concentraion of NaOH = 0.1023 M Volume of acetic acid = 20 ml Volume of NaOH required to reach equivalent point = 20.80ml Calculate the anticipated PH after 9.0ml of the standard NaOH have been added to the acetic acid solution. Show all calculations and explain/justify any assumptions.
if 38.03 ml of 0.08289 M hydrochloric acid requires 21.33 ml of the NaOH solution to...
if 38.03 ml of 0.08289 M hydrochloric acid requires 21.33 ml of the NaOH solution to reach the endpoint, what is the molarity of the NaOH solution?
A 25.00 mL sample of a solution required 21.40 mL of 0.1185 M NaOH to reach...
A 25.00 mL sample of a solution required 21.40 mL of 0.1185 M NaOH to reach the visual endpoint. How many moles of H+ were present in the aliquot?  Give your answer to four decimal places.
a) If your titration solution is 0.409 M in NaOH, and the endpoint occues at 13.00...
a) If your titration solution is 0.409 M in NaOH, and the endpoint occues at 13.00 mL of titrant, how many mmol of NaOH are required to reach the endpoint? ____ mmol NaOH b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH? ____ mmol HC2H3O2 c) How many grams of acetic acid is this? ____ grams HC2H3O2 d) If the mass of analyte is 10.15 grams, what is the mass % of acetic acid...
(a) If your titration solution is 0.410 M in NaOH, and the endpoint occurs at 13.50...
(a) If your titration solution is 0.410 M in NaOH, and the endpoint occurs at 13.50 mL of titrant, how many mmol of NaOH are required to reach the endpoint? __?__mmol NaOH (b) How many mmol of acetic acid (HC2H3O2) are required to react with the NaOH? __?__ mmol HC2H3O2 (c) How many grams of acetic acid is this? __?__ grams HC2H3O2 (d) If the mass of analyte is 10.05 grams, what is the mass % of acetic acid in...
A 22.5 mL sample of an acetic acid solution is titrated with a 0.175M NaOH solution....
A 22.5 mL sample of an acetic acid solution is titrated with a 0.175M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. What was the concentration of acetic acid in the original (22.5mL) sample? What is the pH of the equivalence point? Ka acetic acid= 1.75E-5
If 25.31 mL of 0.0500 M NaOH solution are required to reach the equivalence point of...
If 25.31 mL of 0.0500 M NaOH solution are required to reach the equivalence point of an acid-base titration of 40.0 mL of 7-Up, how many moles of NaOH were required? Based on your answer to the previous question, how many moles of citric acid are in the 40.0 mL of 7-Up?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT