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Dilution, Titration and Redox Problems

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1. Draw me a beaker for each of the following 3 solutions. Show (label) the particles that exist when 1 particle of each dissolves. You don't have to show the water molecules!
H3P04 C12H22011 HCHOO

2. Balance the following redox reaction that occurs in basic solution.
NO2- (aq) + Al (s) ---> NH3 (g) + Al02- (aq)

3. Describe, in detail, how you'd make 100 mL of a 2.0 M HC1 solution from a 6.0 M HC1 solution. How much solvent do you expect to add? For this essay, you may answer in list form (ie, Step 1 ...etc). Support your answer with calculations to indicate amounts of all liquids.

Equipment available: graduated cylinder, pipet, volumetric flask, beaker, 100 niL of 6.OM HC1, 100 mL of distilled water, 100 mL of tap water

4. 100 mL of 0.5M NaOH are used to titrate 25 mL of a diprotic acid (acid with two protons), what was the
concentration of the acid solution?

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https://brainmass.com/chemistry/oxidation-reduction-and-electrochemistry/dilution-titration-redox-problems-148209

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H3PO4 (phosphoric acid) is a strong acid. Strong acids ionize completely. When it dissolves, it will form 1 PO43- ion and 3 H+ ions (if we are talking about one molecule).

Reaction: H3PO4 ----> 3H+ + PO43-

C12H22O11 (sucrose or table sugar) is what is known as a molecular solid. When it dissolves, it does not split up into ions.

HCHOO (formic acid) is a weak acid. If there is a large amount, just a few molecules will dissolve (split into ions). If we are talking about one of these molecule that splits into ions...

Reaction: HCHOO ----> H+ + HCOO-

1. Split into half reactions (group molecules containing the same atom).

NO2-(aq) ----> NH3(g)
Al(s) ----> AlO2-(aq)

2. Add enough H2O to balance O on both sides.

NO2-(aq) ----> NH3(g) + 2 H2O
2 H2O + Al(s) ----> AlO2-(aq)

3. Add H+ to balance H.

7 H+ + NO2-(aq) ----> NH3(g) + 2 H2O
2 H2O + Al(s) ----> AlO2-(aq) + 4 H+

4. Add electrons to balance the charge (electrons have a negative charge).

6 e + 7 H+ + NO2-(aq) ----> NH3(g) + 2 H2O (-6 + 7 + -1 = 0 ----> 0 + 0 = 0 ...

Solution Summary

Detailed explanations are presented for dilution, redox and titration problems.

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