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Chapter 17: Problem 3

What is the common ion effect?

Short Answer

Expert verified
The common ion effect is the shift in equilibrium caused by the addition of an ion already present in the solution, leading to changes in the solubility of salts.

Step by step solution

01

Define the Common Ion Effect

The common ion effect refers to the phenomenon where the addition of an ion that is already present in a solution can affect the equilibrium of the solution. To understand this effect, one should be familiar with the basic principles of chemical equilibrium and Le Chatelier's principle.
02

Explain with Le Chatelier's Principle

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In the context of the common ion effect, adding more of a common ion shifts the equilibrium away from forming more of that ion, often resulting in less soluble salts precipitating or reducing the solubility of an ionic compound.
03

Illustrate with an Example

Consider a saturated solution of calcium fluoride (CaF2), which is in equilibrium with its ions: CaF2(s) 鈬� Ca2+(aq) + 2F鈭�(aq)Adding a soluble fluoride salt, such as sodium fluoride (NaF), which dissociates to provide an additional source of F鈭� ions, will shift the equilibrium to the left, reducing the solubility of calcium fluoride due to the increased concentration of fluoride ions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Imagine a busy market where people come and go, buying and selling goods all day. Chemical equilibrium is not too different; it represents a state where the rate of the forward reaction, where reactants turn into products, matches the rate of the backward reaction, where products revert to reactants. This does not mean the reactants and products stop changing, rather they do so at a constant, unchanging rate, similar to a busy market that stays busy throughout the day.

This balanced state can be illustrated by a simple chemical equation where the reactants and products are represented by letters, such as A and B forming C:

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Most popular questions from this chapter

Consider the titration of a 25.0-mL sample of 0.115 M RbOH with 0.100 M HCl. Determine each quantity. a. the initial pH b. the volume of added acid required to reach the equivalence point c. the pH at 5.0 mL of added acid d. the pH at the equivalence point e. the pH after adding 5.0 mL of acid beyond the equivalence point

Calculate the pH of the solution that results from each mixture. a. 50.0 mL of 0.15 M HCHO2 with 75.0 mL of 0.13 M NaCHO2 b. 125.0 mL of 0.10 M NH3 with 250.0 mL of 0.10 M NH4Cl

Blood is buffered by carbonic acid and the bicarbonate ion.Normal blood plasma is 0.024 M in HCO3- and 0.0012 M H2CO3(pKa1 for H2CO3 at body temperature is 6.1). a. What is the pH of blood plasma? b. If the volume of blood in a normal adult is 5.0 L, what mass of HCl can be neutralized by the buffering system in blood before the pH falls below 7.0 (which would result in death)? c. Given the volume from part b, what mass of NaOH can be neutralized before the pH rises above 7.8?

Suppose that a buffer contains equal amounts of a weak acid and its conjugate base. What happens to the relative amounts of the weak acid and conjugate base when a small amount of strong acid is added to the buffer? What happens when a small amount of strong base is added?

A 25.0-mL volume of a sodium hydroxide solution requires 19.6mL of a 0.189 M hydrochloric acid for neutralization. A 10.0-mL volume of a phosphoric acid solution requires 34.9 mL of the sodium hydroxide solution for complete neutralization. Calculate the concentration of the phosphoric acid solution.
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