Question

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?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

millimoles of NaOH = 8.92 x 0.203 = 1.81

millimoles of HCl = 40 x 0.141 = 5.64

millimoles that actually reacted with antaacid = 5.64 -1.81

                                                                    = 3.83

millimoles of base = 3.83

moles = 3.83 x 10^-3

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
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
A 0.303 g sample of Tums (containing CO2−3CO32- ) is dissolved in 25.0 mL of 0.102...
A 0.303 g sample of Tums (containing CO2−3CO32- ) is dissolved in 25.0 mL of 0.102 M HCl. The hydrochloric acid that is not neutralized by the Tums is back titrated with 8.04 mL of 0.102 M NaOH. How many mol of base are in the antacid?
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?
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 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...
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...
A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water...
A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 16.1 mL of HCl was used to reach the end point, what is the mass % of NaOH in the sample?
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...
A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water...
A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 18.3 mL of HCl was used to reach the end point, what is the mass % of NaOH in the sample? NaOH + HCl --> NaCl + H2O Note: 1. Keep 3 sig figs for your final answer. 2. Answer as percentage.
A .025 mol sample of a weak acid, HA, is dissolved in 485 mL of water...
A .025 mol sample of a weak acid, HA, is dissolved in 485 mL of water and titrated with .41M NaOH. After 29 mL of the NaOH solution has been added the overall pH = 5.414. Calculate the Ka value for HA.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT