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# How can I identify these 4 solutions?

4 pipets contain solutions of Ba(NO3)2, H2SO4, HNO3, and NaOH, although not necessarily in that order. Devise and carry out an experiment to identify the contents of the four pipets and determine the concentration of both acids. The molarity of the NaOH solution is 0.50M.

Please help me? Don't tell me the answers, but at least show me what I'm supposed to do. How do I identify these 4 solutions? I know it has something to do with solubility rules, but I'm just confused

• You want 2 answers... one million) two (3x – five) = eight 6x - 10 = eight (Add 10 to each side) 6x = 18 (Divide each side by means of 6) x = three two) two (3x – five) = -eight 6x - 10 = -eight (Add 10 to each side) 6x = two (Divide each side by means of 6) x = two/6 = one million/three Answer B. (You clear up it as soon as by means of simply getting rid of absolutely the significance signal and substitute with () if wanted. The moment time, difference the signal of the reply at the proper part)
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• Question
How can I identify these 4 solutions?
4 pipets contain solutions of Ba(NO3)2, H2SO4, HNO3, and NaOH, although not necessarily in that order. Devise and carry out an experiment to identify the contents of the four pipets and determine the concentration of both acids. The molarity of the NaOH solution is 0.50M.

Add 5 ml of each solution to 4 test tubes
Use litmus or pH paper to determine which burette contains acid or base.
The litmus paper will be blue in base. The acids will be red.
The pH of acids is less than 7, and the pH of bases is greater than 7.
The test tube that was neutral contains Ba(NO3)2.
The test tubes that were acid contain H2SO4 and HNO3
The test tube that was basic contains NaOH

Add 5 ml of Ba(NO3)2 to 2 test tubes.

To determine the concentration of each of the acids, do a titration with the NaOH.
Titrate acid #1 with the 0.5 M NaOH to determine its normality
Put 75 ml of de-ionized water into a 125 ml flask.
Add 5 drops of Bromthymol blue indicator, the color should be green
Bromthymol blue is yellow in acid, green in neutral, and blue in base.

Add 5 ml of acid #1 to the flask, the color should turn yellow.
Move the flask under the NaOH burette.
Slowly open the stopcock until it produces 1 drop each second.
Keep one hand on the stopcock, so you can close it quickly when needed.
Use the other hand to swirl the burette, to mix the HNO3 with the NaOH and indicator.
Close the stopcock, when the solution turns green. If the solution turns blue, you have added too much NaOH.
In that case, add a few drops of acid #1, until the solution is green. You may have to go back several times to stop at green.

Record the volume of acid #1 and NaOH in a table.
Do this procedure 2 more times.
KEEP 20 ml OF NEUTRALIZED ACID #1 TO BE USED AGAIN.

Use the equation below to determine the normality of the acid #1 solution in each of the 3 trials.
Volume NaOH * 0.5 = volume of acid #1 * Normality of acid #1
Find the average Nomality of acid #1 =

Do the same procedure for acid #2.
Calculate the Normality of acid #2
KEEP 20 ml OF NEUTRALIZED ACID #2 TO BE USED AGAIN.

NOW YOU ARE READY TO USE THE 20 ml OF THE NEUTRALIZED ACIDS.

Pour 10 ml of each acid into separate test tubes.
Add 2 ml of NaOH to the each test tube of acid #1 and acid #2. The color turns blue. (base)
Now the solutions are slightly basic.

Use a pipette to add 1 drop of the Ba(NO3)2 solution to the test tube containing neutralized acid #1.
If the solution in the test tube that gets cloudy the acid is H2SO4
If the solution in the test tube that doesn’t get cloudy the acid is HNO3

Do this same procedure for the test tube of acid#2, to verify your decision concerning which acid is HNO3 or H2SO4

If neither solution gets cloudy, add 2 ml of NaOH, turning the solutions blue.
The neutralized H2SO4 contains Na2SO4 and H2O
The neutralized HNO3 contains NaNO3 and H2O
.
Adding Ba(NO3)2 to Na2SO4 produces BaSO4, which is only slightly soluble in water, but very soluble in acid.
Adding 2 ml of NaOH to the BaSO4 makes the solution more basic and produces Ba(OH)2, which is soluble in water.
Adding drops of H2SO4 to the Ba(OH)2 will produce more BaSO4, but this time in a fro sure basic solution. A precipitate should form. (cloudyness)

The test tube that gets cloudy contains H2SO4, the test tube that doesn’t get cloudy contains HNO3.

Cloudy reaction:
Ba(NO3)2 + Na2SO4 → NaNO3 + BaSO4↓, the down arrow means prepicipitate
BaSO4 is

Please help me? Don't tell me the answers, but at least show me what I'm supposed to do. How do I identify these 4 solutions? I know it has something to do with solubility rules, but I'm just confused
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