#1
30th November 2016, 09:39 AM
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KB Base Ionization Constant
Is there anybody who will provide information about Kb: The base ionization constant? Actually my brother is doing B.Tech and he wants information about Kb: The base ionization constant, so pls give its information as soon as possible.
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#2
30th November 2016, 01:18 PM
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Re: KB Base Ionization Constant
As you are looking for information about Kb: The base ionization constant, so Here I am providing following details: Kb: The base ionization constant Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature. Basic Information 1) Weak bases are less than 100% ionized in solution. 2) Ammonia (formula = NH3) is the most common weak base example used by instructors. The following equation describes the reaction between ammonia and water: NH3 + H2O <==> NH4+ + OH¯ Note that it is an equilibrium condition. The equilibrium constant for this reaction is written as follows: Kc = ( [NH4+] [OH¯] ) / ( [NH3] [H2O] ) However, in pure liquid water, [H2O] is a constant value. To demonstrate this, consider 1000 mL of water with a density of 1.00 g/mL. This 1.00 liter (1000 mL) would weigh 1000 grams. This mass divided by the molecular weight of water (18.0152 g/mol) gives 55.5 moles. The "molarity" of this water would then be 55.5 mol / 1.00 liter or 55.5 M. The solutions studied in introductory chemistry are so dilute that the "concentration" of water is unaffected. So 55.5 molar can be considered to be a constant if the solution is dilute enough. Moving [H2O] to the other side gives: Kc [H2O] = ( [NH4+] [OH¯] ) / [NH3] Since the term Kc [H2O] is a constant, let it be symbolized by Kb, giving: Kb = ( [NH4+] [OH¯] ) / [NH3] This constant, Kb, is called the base ionization constant. It can be determined by experiment and each base has its own unique value. For example, ammonia's value is 1.77 x 10¯5. From the chemical equation above, it can be seen that NH4+ and OH¯ concentrations are in the molar ratio of one-to-one. This will have an important consequence as we move into solving weak acid problems. |
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