Write the Net Ionic Equation for the Precipitation of Manganese(Ii) Carbonate From Aqueous Solution

Assignment 004

Question 1
State whether the following is soluble or insoluble in water:

sodium phosphate

Question 2
Consider the precipitation reactions that could occur when the following pairs of aqueous solutions are mixed:

Reaction 1. silver nitrate + barium chloride

Reaction 2. sodium carbonate + silver sulfate

The formula Ba ( NO₃ )₂ will

A. appear in the net ionic equation for reaction 1 only.

B. appear in the net ionic equation for reaction 2 only.

C. appear in the net ionic equation for both reactions.

D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate

E. not appear in the net ionic equation for either reaction because it is not a correct formula for the any of the ions or precipitates.

Question 3
Consider the precipitation reactions that could occur when the following pairs of aqueous solutions are mixed:

Reaction 1. iron(II) nitrate + sodium carbonate

Reaction 2. calcium nitrate + sodium carbonate

The formula CO₃ ^{2 -} will

A. appear in the net ionic equation for reaction 1 only.

B. appear in the net ionic equation for reaction 2 only.

C. appear in the net ionic equation for both reactions.

D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate

E. not appear in the net ionic equation for either reaction because it is not a correct formula for the any of the ions or precipitates.

Question 4
We obtain a precipitate when we mix aqueous solutions of iron(II) nitrate and ammonium sulfide. The net ionic equation for this reaction is:

**^*+(aq) + *^*-(aq)

***(s)

Question 5
True or False. The net ionic equation for the aqueous reaction between hydrochloric acid and sodium hydroxide is

H ⁺( aq) + OH^- (aq) H₂O(l)

Question 6
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in

MnO₂ Mn^{2 +}

we can balance H and O by adding H₂O and/or H⁺ if the reaction is occurring in an acidic solution. What would be the coefficient of H⁺ when the half reaction above is balanced?

A. 2	B. 4
C. 8	D. 12
E. none of the above

Question 7
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in

NO₃ ^- NO₂ ^-

we can balance H and O by adding H₂O and/or

OH^-

if the reaction is occurring in basic solution. When H and O are balanced in the example above, what would be the coefficient of H₂O?

A. 1	B. 2
C. 4	D. 8
E. none of the above

Question 8
Consider a half reaction where all the atoms are already balanced, as in

SO₂ + H₂O 2 H⁺ + SO₃

When this half reaction is finally balanced, what would be the coefficient of e^-?

A. 1	B. 2
C. 3	D. 4
E. none of the above

Question 9
In the redox reaction:

permanganate ions + chloride ions

chlorine molecules + manganese(II) ions; acidic medium

In the balanced equation with the simplest set of whole number coefficients the sum of all the coefficients is ____ and number of electrons transferred is ____.

A. 12, 2

B. 43 ,10

C. 7 ,2

D. 5, 2

E. 13, 2

Question 10
Balance the following reaction:

����������� * S^{2 - � +} � * Cr₂O₇ ^{2 -} � +� ** H⁺ �� � �� * Cr^{3 +} � +� * S� +� * H₂O

Question 11
Balance the following:

����������� * MnO₄ ^- � + � * H₂O � +� * ClO₂ ^- �� � �� * MnO₂ � +� *

OH^-

� +� * ClO₄ ^- (basic)

Question 12
Balance the following reaction:

����������� * Br₂ � + � * KOH �� � �� * KBrO₃ � +� * KBr � +� * H₂O

ANSWERS

Question 1 (00000001A0503101, Variation No. 70): A. soluble

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See Rule 1 below. Here's a set of solubility rules you can use. Apply them in the order listed -- once you find a rule that applies, IGNORE the rest.

Rule 1. All compounds containing sodium, potassium, ammonium, nitrate, nitrite, perchlorate, or acetate are soluble.
Rule 2. Most chlorides, bromides, and iodides are soluble. Exception: insoluble if the cation is silver, mercury(I), or lead(II).
Rule 3. Most sulfates are soluble. Exception: insoluble if the cation is strontium, barium, lead(II), mercury(I).
Rule 4. Most hydroxides, phosphates, sulfides, chromates, and carbonates are insoluble (see Rule 1 for exceptions; ALSO: group IIA sulfides are soluble, and hydroxides of calcium, strontium, and barium are moderately soluble.)

Imagine a powdered or crystalline sample of the compound comparable in size to about a grain of rice. If you can get this sample to dissolve in about a teaspoon of water, the compound is soluble. Moderately soluble means that you will notice only some of it will dissolve.

Note that insoluble does not mean zero solubility -- it means very, very, very slightly soluble.

Question 2 (00000001A0503109, Variation No. 53): D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate

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Ba(NO₃ )₂ is a soluble salt and will dissociate into ions which are spectator ions in reaction 1. The net ionic equations for a precipitation reaction should only show the ions that reacted and the precipitate formed. The net ionic equations for the two reactions above are:
Ag⁺(aq) + Cl ^-(aq)

AgCl(s)
and
2 Ag⁺(aq) + CO₃ ^{2 -}(aq)

Ag₂CO₃ (s)

Question 3 (00000001A0503110, Variation No. 57): C. appear in the net ionic equation for both reactions.

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The net ionic equation for a precipitation reaction should only show the ions that reacted and the precipitate formed. The net ionic equations for the two reactions above are:
Fe^{2 +}(aq) + CO₃ ^{2 -}(aq)

FeCO₃ (s)
and
Ca^{2 +}(aq) + CO₃ ^{2 -}(aq)

CaCO₃ (s)

Question 4 (00000001A0503301, Variation No. 28): Fe2S2FeS
The completed passage is:
We obtain a precipitate when we mix aqueous solutions of iron(II) nitrate and ammonium sulfide. The net ionic equation for this reaction is:

Fe^{2 +}(aq) + S^2-(aq)

FeS(s)

Feedback
See the solubility rules below to determine which of the two possible products (iron(II) sulfide or ammonium nitrate) is the precipitate. To write down the net ionic equation for a precipitation reaction, simply

1.

write down the formula of the precipitate on the right-hand side (with (s) label)
2. write down the formulas of the cation and anion on the left-hand side; separate the formulas by a + symbol; put (aq) label
3. balance the equation.

Of course,you must first know which pair of cation-anion would precipitate. For this, you need to know your solubility rules. If the cation-anion combination is insoluble in water, then it will give you a precipitate.

Here's a set of solubility rules you can use. Apply them in the order listed -- once you find a rule that applies, IGNORE the rest.

Rule 1. All compounds containing sodium, potassium, ammonium, nitrate, nitrite, perchlorate, or acetate are soluble.
Rule 2. Most chlorides, bromides, and iodides are soluble. Exception: insoluble if the cation is silver, mercury(I), or lead(II).
Rule 3. Most sulfates are soluble. Exception: insoluble if the cation is strontium, barium, lead(II), mercury(I).
Rule 4. Most hydroxides, phosphates, sulfides, chromates, and carbonates are insoluble (see Rule 1 for exceptions; ALSO: group IIA sulfides are soluble, and hydroxides of calcium, strontium, and barium are moderately soluble.)

Note that insoluble does not mean zero solubility -- it means very, very, very slightly soluble. Moderately soluble means that it may or may not precipitate depending on the concentration of the ions.

Question 5 (00000001A0504103, Variation No. 30): A. True

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HCl is a strong acid. NaOH is a strong base. To be able to answer this type of question, you need to know:
1. the six strong acids: HCl,

HBr , HI

, HNO₃ , HClO₄ , and H₂SO₄ . All other acids are weak.

2. the strong, soluble (and moderately soluble) bases -- Group I hydroxides (e.g. NaOH); Group II hydroxides (except Mg(OH)₂ which is insoluble).

When you have a reaction between a strong acid and a strong base, the net ionic equation is just: H ⁺( aq) +

OH^-

(aq)

H₂O.

Question 6 (00000001A0506102, Variation No. 77): B. 4

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Balance O by adding 2 H₂O to the right side. Then balance H by adding 4 H⁺ to the left side.
4 H⁺ + MnO₂

Mn^{2 +}+ 2 H₂O NOTE: we're not done yet. We still have to balance charges. We'll deal with that in another question.

Here's a systematic way to balance H and O in acidic solution (assuming all other atoms have been balanced). Do the following: Step 1. identify the side with fewer O atoms; for every O atom you need, add H₂O to this side. Do this first.
Step 2. After Step 1, identify the side with fewer H atoms. For every H atom you need, add H⁺ to this side.

Example: NO₃ ^-

NO
Step 1. The right side needs TWO more O. Add TWO H₂O to that side. Now we have
NO₃ ^-

NO + 2 H₂O
Step 2. Now the left side needs FOUR hydrogens. Add FOUR H⁺ to the left side. Now we have
4 H⁺ + NO₃ ^-

NO + 2 H₂O

Question 7 (00000001A0506103, Variation No. 1): A. 1

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See example below. NOTE: we're not done yet. We still have to balance charges. We'll deal with that in another question.

Here's a systematic way to balance H and O in basic solution (assuming all other atoms have been balanced). This is NOT the only way. Do the following: Step 1. identify the side with fewer O atoms; for every O atom you need, add H₂O to this side. Do this first.

Step 2. After Step 1, identify the side with MORE H atoms. For every H atom you have in excess, add one

OH^-

to this side AND one H₂O to the other side. This should be easy to remember: excess H +

OH^-

yields H₂O on the other side.

Step 3. Check both sides: if H₂O shows up on both sides, simplify.

Example: NO₃ ^-

NO₂ ^-
Step 1. The right side needs ONE more O. Add ONE H₂O to that side. Now we have
NO₃ ^-

NO₂ ^- + H₂O
Step 2. Now you have TWO excess hydrogens on the right. So we add TWO OH^- to the right side AND TWO H₂O to the left side. Now we have
2 H₂O + NO₃ ^-

NO₂ ^- + H₂O + 2 OH^-
Step 3. Examine both sides: you have 2 H₂O on the left and one on the right. Simplify by subtracting (taking out) one H₂O from each side. So, now we have:
H₂O + NO₃ ^-

NO₂ ^- + 2

OH^-

Question 8 (00000001A0506106, Variation No. 75): B. 2

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Total charges: Left 0, Right +2. Right side is algebraically higher. 2-0 = 2; Add 2 electrons to right side.
SO₂ + H₂O

2 H⁺ + SO₃ + 2e^- Here's a systematic way to balance charges:

Step 1. Add up charges for left and right side

Step 2. Identify the side with the algebraically higher charge (more positive or less negative). Examples:
Left -2 vs. Right -5; left side has algebraically higher charge
Left +1 vs. Right +2; right side has algebraically higher charge
Left +1 vs. Right -2; left side has algebraically higher charge

Step 3. Calculate difference in charges; add this many electrons to side with higher charge in order to make the charges equal.

Example:
14 H⁺ + Cr₂O₇ ^{2 -}

2 Cr^{3 +} + 7 H₂O
Step 1. Total charges:
Left 14x(+1) + 1x(-2) = +12
Right 2x(+3) = +6.
Step 2. Left side charge is algebraically higher.
Step 3. 12-6 = 6; Add 6 electrons to left side. This will bring down total charge on left side to +6, making it equal to the right side charge.

Here's another way (same example above):
Step 1. Identify atoms that have changed oxidation number.
In this example TWO Cr atoms changed from +6 to +3.
Step 2. Calculate total change oxidation number. This will be the number of electrons to add.
In this example, total change is 6 (which is 3 per Cr atom TIMES 2 Cr atoms)
Step 3. If oxidation number went down, add electrons to left side; otherwise add to right side.
In this example, charge went down, so add 6 electrons to left side.
Note: this method (using oxidation number) works even if not all the atoms are balanced, provided that the atoms the have changed oxidation number are balanced.

Therefore:
6e^- + 14 H⁺ + Cr₂O₇ ^{2 -}

2 Cr^{3 +} + 7 H₂O

Question 9 (00000001A0506108, Variation No. 81): B. 43 ,10

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Half reactions:
MnO₄ ^-

Mn^{2 +}
Cl ^-

Cl₂
The overall equation is:
16H⁺ + 2MnO₄ ^- + 10Cl ^-

2Mn^{2 +} + 5Cl₂ + 8H₂O Here are the general steps in balancing a redox reaction.
Step 1. split the reaction into two half reactions
Step 2. balance each half reaction (the ones shown above are not balanced)
Step 3. multiply the half reactions in order to balance electrons lost and electrons gained;
Step 4. then combine the two half reactions from step 3.
Step 5. simplify coefficients if necessary.

Question 10 (00000001A0506306, Variation No. 70): 3114237
The completed passage is:
Balance the following reaction:

����������� 3 S^2- � + � 1 Cr₂O₇ ^{2 -} � +� 14 H⁺ �� � �� 2 Cr^{3 +} � +� 3 S� +� 7 H₂O

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S^{2 - � +} � Cr₂O₇ ^{2 -} � +� H⁺ �� � �� S� +� Cr^{3 +} � +� H₂O

Red ���� Cr₂O₇ ^{2 -} � + � 14 H⁺ � +� 6 e^- �� � �� 2 Cr^{3 +} � +� 7 H₂O

Ox������ 3 (S^2- �� � �� S� + � 2 e^-)

Bal������ 3 S^{2 - � +} � Cr₂O₇ ^{2 -} � +� 14 H⁺ �� � �� 2 Cr^{3 +} � +� 3 S� +� 7 H₂O

Question 11 (00000001A0506308, Variation No. 4): 423443
The completed passage is:
Balance the following:

����������� 4 MnO₄ ^- � + � 2 H₂O � +� 3 ClO₂ ^- �� � �� 4 MnO₂ � +� 4 OH^- � +� 3 ClO₄ ^- (basic)

Feedback

ClO₂ ^- � + � MnO₄ ^- �� � �� MnO₂ � +� ClO₄ ^- ���������� (basic)

Red����� 4 (MnO₄ ^- � + � 2 H₂O � +� 3 e^- �� � �� MnO₂ � +� 4 OH^- )

Ox������ 3 (ClO₂ ^- � + � 4 OH^- �� � �� ClO₄ ^- � +� 2 H₂O � 4 e^-)

Bal������ 4 MnO₄ ^- � + � 2 H₂O � +� 3 ClO₂ ^- �� � �� 4 MnO₂ � +� 4 OH^- �+� 3 ClO₄ ^-

Question 12 (00000001A0506310, Variation No. 41): 36153
The completed passage is:
Balance the following reaction:

����������� 3 Br₂ � +� 6 KOH �� � �� 1 KBrO₃ �+� 5 KBr � +� 3 H₂O

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Br₂ � + � KOH �� � �� KBrO₃ � +� KBr � +� H₂O

Net����� Br₂ � + � OH^- �� � �� BrO₃ ^- � +� Br^- � +� H₂O

Red����� 5 ( Br₂ � + � 2 e^- �� � �� 2 Br^-)

Ox������ Br₂ � + � 12 OH^- �� � �� 2 BrO₃ -� +� 6 H₂O � + 10 e^-

Net Bal����������� 6 Br₂ � + � 12 OH^- �� � �� 2 BrO₃ ^- � +� 10 Br^- � +� 6 H₂O

Bal������ 6 Br₂ � + � 12 KOH �� � �� 2 KBrO₃ � +� 10 KBr � +� 6 H₂O (common factor 2 - must reduce)

����������� 3 Br₂ � +� 6 KOH �� � �� KBrO₃ � +� 5 KBr � +� 3 H₂O

Write the Net Ionic Equation for the Precipitation of Manganese(Ii) Carbonate From Aqueous Solution

Source: http://www.faculty.sfasu.edu/janusama/Assignment004.htm

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