Examples of ion product constant for water in the following topics:
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- The value of the dissociation constant of water, KW, is $1.0\times 10^{-14}$.
- An acid dissociation constant, Ka, is the equilibrium constant for the dissociation of an acid in aqueous solution.
- For a generalized reaction of a base, B, in water, we have:
- A water molecule is protonated to form a hydronium ion in the process.
- A water molecule protonates a neighboring water molecule, yielding hydronium and hydroxide ions.
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- The solubility product constant (Ksp) is the equilibrium constant for a solid that dissolves in an aqueous solution.
- All of the rules for determining equilibrium constants continue to apply.
- For substances in which the ions are not in a 1:1 ratio, the stoichiometric coefficients of the reaction become the exponents for the ions in the solubility-product expression:
- At a certain temperature, the solubility of Fe(OH)2 in water is 7.7 x 10-6 mol/L (M).
- The solubility product constants of a number of substances.
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- In equilibrium, the electrochemical potential will be constant everywhere for each species.
- For example, if a glass of water has sodium ions (Na+) dissolved uniformly in it, and an electric field is applied across the water, then the sodium ions will tend to get pulled by the electric field toward one side.
- Likewise, if a glass of water has a lot of dissolved sugar on one side and none on the other side, each sugar molecule will randomly diffuse around the water until there is and equal concentration of sugar everywhere.
- This is a particularly high equilibrium constant value, indicating that the equilibrium strongly favors the formation of products (the reaction is effectively irreversible towards the formation of products).
- Calculate the equilibrium constant, K, for a galvanic cell using the Nernst equation
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- Historically, the equilibrium constant Kb for a base has been defined as the association constant for protonation of the base, B, to form the conjugate acid, HB+.
- As with any equilibrium constant for a reversible reaction, the expression for Kb takes the following form:
- Recall that in water, the concentration of the hydroxide ion, [OH−], is related to the concentration of the hydrogen ion by the autoionization constant of water:
- The general equation for a base dissociation constant, where B is the base, HB is its conjugate acid, and OH- is hydroxide ions.
- Calculate the Kw (water dissociation constant) using the following equation: Kw = [H+] x [OH−] and manpulate the formula to determine [OH−] = Kw/[H+] or [H+]=Kw/[OH-]
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- For example, table salt (NaCl) placed in water eventually dissolves.
- This is because the rate of the forward (reactant to product) and reverse (product to reactant) reactions are equal.
- For example, this would be like trying to dissolve solid table salt (NaCl) in a solution where the chloride ion (Cl-) is already present.
- For a simple dissolution process, the addition of more of one of the ions (A+) from another compound will shift the composition to the left, reducing the concentration of the other ion (B-), effectively reducing the solubility of the solid (AB).
- For example, when calcium fluoride dissolves into calcium and fluoride ions, the solubility product expression is:
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- Hydrogen ions are spontaneously generated in pure water by the dissociation (ionization) of a small percentage of water molecules into equal numbers of hydrogen (H+) ions and hydroxide (OH-) ions.
- The hydroxide ions remain in solution because of their hydrogen bonds with other water molecules; the hydrogen ions, consisting of naked protons, are immediately attracted to un-ionized water molecules and form hydronium ions (H30+).
- By convention, scientists refer to hydrogen ions and their concentration as if they were free in this state in liquid water.
- The concentration of hydrogen ions dissociating from pure water is 1 × 10-7 moles H+ ions per liter of water.
- Maintaining a constant blood pH is critical to a person's well-being.
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- As the equilibrium constant approaches infinity, the reaction tends to form 100 percent products.
- The equilibrium constant K = 1 states that there will be 50 percent products and 50 percent reactants.
- Because the equilibrium constant is used for calculating the concentrations of weak acids, very little water actually reacts relative to the total concentration.
- The concentration of water during the reaction is, therefore, a constant, and can be excluded from the expression for K.
- A water molecule is protonated to form a hydronium ion in the process.
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- Solubility product principle is used in qualitative analysis to determine composition of a compound by separation of ions in a solution.
- Solubility-product constants can be used to devise methods for separating ions in a solution by selective precipitation.
- The entire traditional qualitative-analysis scheme is based on the use of these equilibrium constants to determine the correct precipitating ions and the correct strategy.
- Because cationic analysis is based on the solubility products of the ions, meaningful results can be obtained only if separation is performed in a specified sequence.
- For example, both Ba2+ and Sr2+ will react with the SO42- ion to form a solid.
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- The result is the formation of a hydroxide ion (OH-) and a hydronium ion (H3O+).
- Because this is a special equilibrium constant, specific to the self-ionization of water, it is denoted KW; it has a value of 1.0 x 10−14.
- If we plug in the above value into our equation for pH, we find that:
- The self-ionization of water produces hydronium and hydroxide ions in solution.
- Explanation of self-ionization of water and the formation of hydronium and hydroxide ions.
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- This is because Le Chatelier's principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product.
- Scientists take advantage of this property when purifying water.
- In areas where water sources are high in chalk or limestone, drinking water contains excess calcium carbonate CaCO3.
- With such a small solubility product for CaF2, you can predict its solubility << 0.10 moles per liter.
- If our prediction is valid, we can simplify the solubility-product equation: