Question

a. The concentration of HCl in a bottle was determined as follows: 4.00 mL of the...

a. The concentration of HCl in a bottle was determined as follows: 4.00 mL of the acid were diluted to 150.00 mL 28.38 mL of the diluted acid were titrated with 0.08570-M NaOH 41.07 mL of the base were required to reach the end point What is the HCl concentration in the bottle?

b. The following successive dilutions are applied to a stock solution that is 4.70 M sucrose Solution A = 58.0 mL of the stock sollution is diluted to 288 mL Solution B = 144.0 mL of Solution A is diluted to 334 mL Solution C = 244.8 mL of Solution B is diluted to 495 mL What is the concentration of sucrose in solution C?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

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
HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) 4.00 mL of an unknown...
HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) 4.00 mL of an unknown acid solution containing HCl was added to flask containing 30.00 mL of deionized water and two drops of an indicator. The solution was mixed and then titrated with NaOH until the end point was reached. Use the following titration data to determine the mass percent of HCl in the acid solution. Mass of flask: 125.59 g Mass of flask + acid solution: 129.50 g...
1) How many moles of acid are there in 17 ml of a 7.1 M HCl...
1) How many moles of acid are there in 17 ml of a 7.1 M HCl solution? 2) 18 ml of a 1.0 M HCl solution was added to a flask and titrated against a solution of 3.8 M NaOH. How many ml of the NaOH solution was required to neutralize the acid? 3) What volume (in liters) of a 2.8 M NaOH solution would neutralize 4.8 moles of H2SO4 (diprotic)? 4) Using the formula M1V1 = M2V2, calculate the...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate the concentration of the acid in the original solution, the pH of the original HCl solution and the original NaOH solution
Using the concentration values determined in this investigation for M(OH)2 and HCl, calculate the pH for...
Using the concentration values determined in this investigation for M(OH)2 and HCl, calculate the pH for a strong acid + strong base titration in which 5.00 mL of the M(OH)2 was transferred via pipet to a beaker and HCl was added from the buret. Concentration of base is 0.07 M Concentration of acid is 0.0532 M Calculate the pH: a) before any HCl is added b) after the addition of 4.00 mL HCl c) after the addition of 9.00 mL...
Calculate the molar concentration (molarity) of the HCl soluition and the HC2H3O2 solution using the following...
Calculate the molar concentration (molarity) of the HCl soluition and the HC2H3O2 solution using the following information: 50mL of distilled water were mixed with 10 mL of the acid. A titration was performed using 0.100 M of NaOH solution. Find the average molar concentration for the two trials. HCl: Equivalance point: 11.17 mL of NaOH added. Acetic Acid: Equivalence point: 10.26 mL of NaOH added.
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl....
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH after 7.6 mL of acid are added ______ Find the pH after 20 mL of acid are added _______
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl....
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH after 3.1 mL of acid are added. Find the pH after 20 mL of acid are added.
Calculate the concentration of a solution of HCl if a 25.00 mL sample required 24.87 mL...
Calculate the concentration of a solution of HCl if a 25.00 mL sample required 24.87 mL of 0.1014 M NaOH to titrate to the endpoint.
A student pipetted 25.00 mL of a stock solution that was 0.1063 M HCl into a...
A student pipetted 25.00 mL of a stock solution that was 0.1063 M HCl into a 100.00 mL volumetric flask and diluted the solution to the volumetric flask calibration mark with deionized water. Calculate the concentration of the diluted solution.
An antacid tablet was dissolved in 26.00 mL of 0.650 M HCl. The excess acid was...
An antacid tablet was dissolved in 26.00 mL of 0.650 M HCl. The excess acid was back-titrated with exactly 11.34 mL of 1.05 M NaOH. The average weight of a tablet is 0.834 g. The tablet came from a bottle of 150 tablets that cost $3.99. Calculate the moles of HCl neutralized by the tablet Calculate the mass effectiveness of the antacid Calculate the cost-effectiveness of the antacid
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT