Question

concentration HCl in mmol/mL= 0.403 concentration NaOH in mmol/mL= 0.11 Part IIC. Scoring Scheme: 3-3-2-1. Based...

concentration HCl in mmol/mL= 0.403

concentration NaOH in mmol/mL= 0.11

Part IIC.
Scoring Scheme: 3-3-2-1.
Based upon the molarities of the standardized HCl and standardized NaOH, and the volumes of each added to analyze the antacid samples, calculate the MILLIMOLES of HCl needed to neutralize the tablet per MILLIGRAM of sample.
Use the same sequence as you entered the mass of the samples and report your results to 4 significant figures.

Entry # mass tablet(g) mass antacid(g) Vol HCl(mL) Vol NaOH(mL) mmoles HCl/mg
#1: 0.4140 1.384 22.20 22.20
#2: 0.4220 1.380 20.50 20.50
#3: 0.4140 1.422 21.10 21.10

Homework Answers

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
1. If the HCl used in this experiment was prepared at a concentration of 0.15M. Using...
1. If the HCl used in this experiment was prepared at a concentration of 0.15M. Using your volume of HCl neutralized by your whole tablet, calculate the mass of active ingredient (CaCO3) in the tablet in milligrams. 2. Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196g sample of antacid containing an unknown amount of triprotic base Al(OH)3 was reacted with 25.0mL of 0.111M HCl. The resulting solution was then...
Aspirin Determination using Back Titration Began with 0.1 M HCl solution and 0.1 M NaOH solution....
Aspirin Determination using Back Titration Began with 0.1 M HCl solution and 0.1 M NaOH solution. 1:2 aspirin to NaOH mole ratio, therefore, mol HCl (back titration) = excess mol NaOH AND total mol NaOH used - excess mol NaOH=required mol NaOH AND required mol NaOH (1 mol aspirin / 2 mol NaOH) = mol aspirin. Each step builds off the last 1) Determination of HCl Concentration. This was the first titration. Concentrated HCl was diluted to .1M before titrating....
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...
In Part 1 of Lab 2 you will make and standardize a solution of NaOH(aq). Suppose...
In Part 1 of Lab 2 you will make and standardize a solution of NaOH(aq). Suppose in the lab you measure the solid NaOH and dissolve it into 100.0 mL of water. You then measure 0.2013 g of KHP (KC8H5O4, 204.22 g/mol) and place it in a clean, dry 100-mL beaker, and then dissolve the KHP in about 25 mL of water and add a couple of drops of phenolphthalein indicator. You titrate this with your NaOH(aq) solution and find...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT