Question

0.5358 grams of liquid antacid is mixed with 50 mL of DI water. An appropriate indicator...

0.5358 grams of liquid antacid is mixed with 50 mL of DI water. An appropriate indicator and 180.00 mL of the HCl solution are added. If it requires 120.31 mL of the KOH solution to back titrate the excess HCl to the endpoint, how many moles of acid were neutralized by the antacid sample?

__________ moles

Homework Answers

Answer #1

Assume that the moalrity of KOH and HCl is 0.100 M

The reaction of KOH and HCl ia as follows:

KOH+ HCl = KCl + H2O

First calculate the moles of KOH = Molarity * volume in L

= 0.100 *120.31/1000

= 0.012 Moles of KOH reacted with the same mole of HCl

Total mole of HCl = 0.100 *180/1000

= 0.0180 moles

Moles of HCl which are reacted with antacid = total moles – moles which are reacted KOH

= 0.0180 -0.012

= 0.006 Mole HCl

0.006 moles of acid were neutralized by the antacid sample. If the moalrity of KOH and HCl is 0.100 M

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
10.94 mL of KOH solution is required to titrate 0.5361 g of potassium hydrogen phthalate (KHP;...
10.94 mL of KOH solution is required to titrate 0.5361 g of potassium hydrogen phthalate (KHP; FW=204.224) to the Phenolphthalein endpoint. If 27.96 mL of this solution is required to titrate 96.34 mL of HCl to the Phenolphthalein endpoint, what is the concentration of the HCl solution? __________ M One antacid tablet is ground and dissolved in 50 mL of DI water. Indicator and 100.00 mL of the HCl solution are added. If it requires 23.22 mL of the KOH...
Suppose a student adds 25.00 mL of 1.041 M HCl to a 1.50 g antacid tablet....
Suppose a student adds 25.00 mL of 1.041 M HCl to a 1.50 g antacid tablet. The student boils and then titrates the resulting solution to the endpoint with 0.4989 M NaOH. The titration requires 21.1 mL NaOH to reach the endpoint. How many moles of HCl were neutralized by the NaOH?How many moles of HCl were neutralized by the tablet?
A 0.352g sample of an antacid is dissolved in water. To it is added 40.00ml of...
A 0.352g sample of an antacid is dissolved in water. To it is added 40.00ml of 0.141M HCl solution? The solution is heated to drive off any CO2 gas. It is then back titrated to endpoint with 8.92ml of a 0.203M solution of NaOH. How many moles of base were in the original sample of the antacid. How many millimoles?
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
A student dissolves 0.326 g of a powdered antacid in 32.36 mL of 0.1034 M HCl....
A student dissolves 0.326 g of a powdered antacid in 32.36 mL of 0.1034 M HCl. The student boils the mixture and then allows it to cool. Lastly, the student adds phenolphthalein indicator to the mixture. 1) Calculate the total number of moles of H+ added to the antacid. 2)Suppose 11.72 mL of 0.1506 M NaOH is required to turn the solution from colorless to pale pink. Calculate the total moles of OH- added. 3) Calculate the difference between total...
0.82 g of Sodium Acetate (NaAc) is filled with DI water up to 100 mL. 2.3...
0.82 g of Sodium Acetate (NaAc) is filled with DI water up to 100 mL. 2.3 mL Acetic Acid is also filled with DI water up to 100 mL. 47.5 mL HAc (0.10 M) and 2.5 mL NaAc (0.10M) are mixed. a) What is the pH of the solution after HAc and NaAc are mixed? b) What is the pH value if 2.0 mL NaOH (0.10M) is added to the solution?
A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of...
A 0.4016 g sample of impure sodium carbonate (soda ash) is dissolved in 50 mL of distilled water. Phenolphthalein was added and 11.40 mL of 0.09942 M HCL was required to reach the first end point. Excess volume of HCL was added according to the lab procedure (if x is the amount of acid needed to complete the first titration, you will add a total of 2x + 10mL of acid to ensure excess). After boiling, the excess acid was...
A 1.379-g sample of commercial KOH contaminated K2CO3 by was dissolved in water, and the resulting...
A 1.379-g sample of commercial KOH contaminated K2CO3 by was dissolved in water, and the resulting solution was diluted to 500.0 mL. A 50.00-mL aliquot of this solution was treated with 40.00 mL of 0.05304 M HCl and boiled to remove CO2. The excess acid consumed 4.55 mL of 0.04925 M (phenolphthalein indicator). An excess of neutral BaCl2 was added to another 50.00-mL aliquot to precipitate the carbonate as BaCO3. The solution was then titrated with 28.56 mL of the...
If you had added 50 mL of water to dissolve the oxalic acid dihydrate instead of...
If you had added 50 mL of water to dissolve the oxalic acid dihydrate instead of 30 mL, would that require more, less, or the same amount of NaOH solution to titrate? Explain
Part 1: What is the concentration of your NaOH(aq) solution? 1.00×10-1 M NaOH Ok. Part 2:...
Part 1: What is the concentration of your NaOH(aq) solution? 1.00×10-1 M NaOH Ok. Part 2: You will then make a Kool-Aid solution by dissolving the powder in an entire packet of lemon-lime Kool-Aid in 250.00 mL solution.  You find the powder in a packet of lemon-lime Kool-Aid has a mass of 3.654 g. In Part 2 of the lab you titrate 5.00 mL of the lemon-lime Kool-Aid (with 5 drops of thymol blue indicator) with your NaOH(aq) solution. The titration...