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Chapter 16: Problem 99

\(\mathrm{Cd}(\mathrm{OH})_{2}\) is an insoluble compound. It dissolves in excess \(\mathrm{NaOH}\) in solution. Write a balanced ionic equation for this reaction. What type of reaction is this?

Short Answer

Expert verified
The balanced ionic equation for the reaction is \(Cd^{2+} + 4OH^- \rightarrow Cd(OH)_4^{2-}\). The reaction is a complexation reaction and a specific type of precipitation reaction.

Step by step solution

01

Writing the balanced molecular equation

First, we will write out the balanced molecular equation for the reaction. We start with the given reactants: Cd(OH)2 and NaOH, and figure out the product. In this reaction, a complex ion is formed, and the equation looks like the following: Cd(OH)2 + 2NaOH → Na2Cd(OH)4
02

Writing the balanced ionic equation

The balanced ionic equation presents the reaction where the reactants and products are shown as ions if they are ionic (dissociate into ions). Let's indicate the states of the substances: Cd(OH)2 (s) + 2NaOH (aq) → Na2Cd(OH)4 (aq). It can be rewritten as the following ionic equation: Cd2+ + 2OH- + 2Na+ + 2OH- → 2Na+ + Cd(OH)42-
03

Identifying the type of reaction

The reaction is a type of 'precipitation reaction' - reactions that involve the formation of an insoluble product (precipitate) when two salt solutions are combined. However, this reaction differs slightly because the precipitate, Cd(OH)2, undergoes a further reaction in the presence of excess NaOH to form a soluble complex ion, Cd(OH)42-. Therefore, it is also considered a complexation reaction.
04

Final Balanced Ionic Equation

Finally, we simplify the equation by removing 'spectator ions'- ions that don't participate in the reaction. Here, Na+ is a spectator ion. After removing it, the final balanced ionic equation becomes: Cd2+ + 4OH- → Cd(OH)42-

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Most popular questions from this chapter

In a titration experiment, \(20.4 \mathrm{~mL}\) of \(0.883 \mathrm{M}\) HCOOH neutralize \(19.3 \mathrm{~mL}\) of \(\mathrm{Ba}(\mathrm{OH})_{2} .\) What is the concentration of the \(\mathrm{Ba}(\mathrm{OH})_{2}\) solution?

A student titrates an unknown monoprotic acid with a \(\mathrm{NaOH}\) solution from a buret. After the addition of \(12.35 \mathrm{~mL}\) of \(\mathrm{NaOH},\) the \(\mathrm{pH}\) of the solution read 5.22. The equivalence point is reached at \(24.70 \mathrm{~mL}\) of \(\mathrm{NaOH}\). What is the \(K_{\mathrm{a}}\) of the acid?

Which of the following has the greatest buffer capacity: (a) \(0.40 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa} / 0.20 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) \(\begin{array}{llll}\text { (b) } 0.40 & M \text { CH }_{3} \text { COONa/0.60 } & \text { M CH }_{3} \text { COOH. }\end{array}\) (c) \(0.30 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa} / 0.60 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}^{2}\)

(a) Assuming complete dissociation and no ionpair formation, calculate the freezing point of a \(0.50 \mathrm{~m}\) NaI solution. (b) What is the freezing point after the addition of sufficient \(\mathrm{HgI}_{2},\) an insoluble compound, to the solution to react with all the free \(\mathrm{I}^{-}\) ions in solution? Assume volume to remain constant.

A \(5.00-\mathrm{g}\) quantity of a diprotic acid was dissolved in water and made up to exactly \(250 \mathrm{~mL}\). Calculate the molar mass of the acid if \(25.0 \mathrm{~mL}\) of this solution required \(11.1 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) KOH for neutralization. Assume that both protons of the acid were titrated.
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