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EC number: 231-439-8 | CAS number: 7550-35-8
Dissociation constant
Administrative data
Link to relevant study record(s)
- Endpoint:
- dissociation constant
- Type of information:
- calculation (if not (Q)SAR)
- Adequacy of study:
- key study
- Study period:
- 2012-07-02
- Reliability:
- 1 (reliable without restriction)
- Qualifier:
- no guideline followed
- Principles of method if other than guideline:
- Calculation based on the water solubility
- Dissociating properties:
- yes
- No.:
- #1
- pKa:
- ca. 2.64
- Temp.:
- 25 °C
- Remarks on result:
- other: Calculation basis: study result on water solubility (no pKa but pKsp; sp = solubility product)
- Conclusions:
- Ksp = 434.33 mol^2/L^2. The log10 of the Ksp value is 2.64.
- Executive summary:
The solubility equilibrium of lithium bromide exists when the chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The equilibrium is an example of dynamic equilibrium in that some individual molecules migrate between the solid and solution phases such that the rates of dissolution and precipitation are equal to one another. When equilibrium is established, the solution is saturated. The concentration of the solute in a saturated solution is known as the solubility. Dissolution with dissociation is a characteristic of salts like lithium bromide. Thus, one type of solubility equilibrium is the reversible dissolution with dissociation. Lithium bromide dissociates in its constituent ions when it is dissolving in water:
LiBr(s) <=> Li+(aq) + Br-(aq)
The corresponding solubility product Ksp is expressed as:
Ksp= [Li+]*[Br-]
The relation between the solubility S and the solubility product Ksp of a salt AmBn like lithium bromide is as follows:
S = (Ksp / (m^m * n^n))^(1/(m+n))
m = 1
n = 1
The solubility S of lithium bromide at 25 degrees Celsius is 1810 g/L and the solubility product Ksp was calculated to be 434.33 mol2/L2. Expressed in a logarithmic form the Ksp is log Ksp= 2.64.
Reference
Lithium bromide dissociates in its constituent ions when it is dissolving in water:
LiBr(s) <=> Li+(aq) + Br-(aq)
The solubility equilibrium of lithium bromide exists when the chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The equilibrium is an example of dynamic equilibrium in that some individual molecules migrate between the solid and solution phases such that the rates of dissolution and precipitation are equal to one another. When equilibrium is established, the solution is saturated. The concentration of the solute in a saturated solution is known as the solubility. Dissolution with dissociation is a characteristic of salts like lithium bromide. Thus, one type of solubility equilibrium is the reversible dissolution with dissociation.
The corresponding solubility product Ksp is expressed as:
Ksp= [Li+]*[Br-]
The relation between the solubility S and the solubility product Ksp of a salt AmBn like lithium bromide is as follows:
S = (Ksp / (m^m * n^n))^(1/(m+n))
The solubility S of lithium bromide at 25 degrees Celsius is 1810 g/L and the solubility product Ksp was calculated to be 434.33 mol2/L2. Expressed in a logarithmic form the Ksp is log Ksp= 2.64.
Description of key information
Ksp was calculated to be 434.33 mol^2/L^2. The log10 of the Ksp value is 2.64.
Key value for chemical safety assessment
- pKa at 20°C:
- 2.64
Additional information
The solubility equilibrium of lithium bromide exists when the chemical compound in the solid state is in chemical equilibrium with a solution of that compound. The equilibrium is an example of dynamic equilibrium in that some individual molecules migrate between the solid and solution phases such that the rates of dissolution and precipitation are equal to one another. When equilibrium is established, the solution is saturated. The concentration of the solute in a saturated solution is known as the solubility. Dissolution with dissociation is a characteristic of salts like lithium bromide. Thus, one type of solubility equilibrium is the reversible dissolution with dissociation. Lithium bromide dissociates in its constituent ions when it is dissolving in water:
LiBr(s) <=> Li+(aq) + Br-(aq)
The corresponding solubility product Ksp is expressed as:
Ksp= [Li+]*[Br-]
The relation between the solubility S and the solubility product Ksp of a salt AmBn like lithium bromide is as follows:
S = (Ksp / (m^m * n^n))^(1/(m+n))
m = 1
n = 1
The solubility S of lithium bromide at 25 degrees Celsius is 1810 g/L and the solubility product Ksp was calculated to be 434.33 mol2/L2. Expressed in a logarithmic form the Ksp is log Ksp= 2.64.
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