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# Write complete ionic and net ionic equations for each of

ISBN: 9780321910295 34

## Solution for problem 78P Chapter 7

Introductory Chemistry | 5th Edition

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Problem 78P

PROBLEM 78P

Write complete ionic and net ionic equations for each of the reactions in Problem.

Problem

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of solutions is mixed. If no reaction occurs, write NO REACTION.

(a) potassium chloride and lead(II) acetate

(b) lithium sulfate and strontium chloride

(c) potassium bromide and calcium sulfide

(d) chromium(III) nitrate and potassium phosphate

Step-by-Step Solution:

Solution 78P

Here, we are going to write complete ionic and net ionic equations for the given reactions.

Step 1:

Net ionic equation shows only those chemical species which are taking part in a chemical reaction. while writing the ionic equation, we write out all the soluble compounds as ions and eliminate ions common to both the reactants and products. This will give us the net ionic equation.

Step 2:

When potassium chloride and lead(II) acetate are mixed, potassium acetate and lead chloride are formed. The balanced molecular equation for the reaction is:

2KCl(aq) + Pb(CH3COO)2(aq) → 2CH3COOK(aq) + PbCl2(s)↓

Converting the soluble compounds into ions, we get the complete ionic equation.Thus, the complete ionic equation for the above reaction is:

2K+(aq) + 2Cl-(aq) + Pb2+(aq) + 2CH3COO-(aq) → PbCl2(s)↓ + 2K+(aq) + 2CH3COO-(aq)

Here, K+ and CH3COO- behaves as spectator ion(ion existing in the same form in both side of the reaction.) and can be eliminated. Thus, the net ionic equation is:

Pb2+(aq) + 2Cl-(aq) → PbCl2(s)↓

Step 3:

b)        When lithium sulfate and strontium chloride are mixed, lithium chloride and strontium sulfate are formed. The balanced molecular equation for the reaction is:

Li2SO4(aq) + SrCl2(aq) → 2LiCl(aq) + SrSO4(s)↓

Converting the soluble compounds into ions, we get the complete ionic equation.Thus, the complete ionic equation for the above reaction is:

2Li+(aq) + SO42-(aq) + Sr2+(aq) + 2Cl-(aq) → SrSO4(s)↓ + 2Li+(aq) + 2Cl-(aq)

Here, Li+ and Cl- behaves as spectator ion and can be eliminated. Thus, the net ionic equation is:

Sr2+(aq) + SO42-(aq) → SrSO4(s)↓

Step 4 of 5

Step 5 of 5

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Write complete ionic and net ionic equations for each of

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