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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 80e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 80e

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# Solved: Write balanced complete ionic and net ionic ISBN: 9780321809247 1

## Solution for problem 80E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Problem 80E

Problem 80E

Write balanced complete ionic and net ionic equations for each reaction. Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass of NaOH needed to precipitate the cadmium ions from 35.0 mL of 0.500 M cadmium nitrate solution. Step1: The reaction between NaOH and Cd(NO ) takes pl3 2n the following manner. 2NaOH(aq) + Cd(NO ) (aq3 2----> Cd(OH) (aq) + 2aNO (aq) 3 In the above reaction, we have seen that 1 mole of Cd(NO ) require3 2oles of NaOH to precipitate out all the cadmium ions. Step2: Now, let us calculate the number of moles of 35.0 ml of 0.500 M Cd(NO ) solution. 3 2 Number of moles of Cd(NO ) solution = Volume of solution in litres X Molarity 3 2 = (35/1000 L) X 0.500 M = 0.035 L X 0.500 M = 0.0175 mol Thus, 0.0175 mol of Cd(NO ) sol3 2 will require (2 x 0.0175 = 0.035) mol of NaOH to precipitate out all the cadmium ions. Step3: Now, 1 mole of NaOH = molar mass of NaOH Therefore, 0.035 mol of NaOH = 0.035 X molar mass of NaOH = 0.035 x 40.01 g [molar mass of NaOH = 40.01 g/mol] = 1.40 g Thus, 1.40 g of NaOH is needed to precipitate the cadmium ions from 35.0 ml of 0.500 M Cd(NO ) solution. 3 2

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##### ISBN: 9780321809247

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