How to calculate Kb from Ka?  (Kb from Ka)
Weak acids and bases dissociate to a small extent in an aqueous solution. A weak acid (HA) breaks down in the water to give H^{+} ions, while a weak base (B) dissociates to produce OH^{–} ions.
The extent of ionization of a weak acid or base in an aqueous solution can be determined by using ionization constants, i.e., K_{a} and K_{b}.
What is K_{a}?
K_{a} stands for acid dissociation constant.
The ionization equilibrium for the dissociation of a weak acid (HA) in an aqueous solution is represented as follows:
The acid dissociation constant (K_{a}) for the above reaction can be represented as equation (i).
Ka = \frac{[H_{3}O^{+}][A^{}]}{[HA][H_{2}O]}………. Equation (i)
Where;
 [H_{3}O^{+}] = concentration of hydronium ions formed in the aqueous solution
 [A^{–}] = concentration of conjugate base of the acid
 [HA] = acid concentration at equilibrium
 [H_{2}O] = concentration of water
As water concentration stays constant throughout the reaction, while [H_{3}O^{+}] = [H^{+}], i.e., the concentration of H^{+} ions released in the aqueous solution. So, equation (i) can be rearranged as equation (ii).
Ka = \frac{[H^{+}][A^{}]}{[HA]}………. Equation (ii)
The greater the strength of an acid, the higher the K_{a} value for its aqueous solution and vice versa.
What is K_{b}?
K_{b} represents the base dissociation constant.
The ionization equilibrium for the dissociation of a weak base (B) in an aqueous solution is represented as:
The base dissociation constant (K_{b}) for the above reaction can be represented as equation (iii).
Kb = \frac{[BH^{+}][OH^{}]}{[B][H_{2}O]}………. Equation (iii)
Where;
 [BH^{+}] = concentration of conjugate acid of the base
 [OH^{–}] = hydroxide ion concentration in aqueous solution
 [B]= concentration of base at equilibrium
Considering the water concentration [H_{2}O] constant, equation (iii) can be rearranged as shown below.
Kb = \frac{[BH^{+}][OH^{}]}{[B]}………. Equation (iv)
The greater the strength of a base, the higher the K_{b} value for its aqueous solution.
What is the relationship between K_{a} and K_{b}?
K_{a} and K_{b} can be interconverted using another chemical entity called the water dissociation constant (K_{w}), as shown in equation (v).
K_{a}. K_{b} = K_{w}………. Equation (v)
⇒ K_{w}= [H^{+}] [OH^{–}]
The value of K_{w} is fixed at 25^{°}C, i.e., room temperature. K_{w} = 1.0 x 10^{14}.
As per equation (v), if the value of K_{a} is known, we can easily determine the value of K_{b} by making K_{b} the subject of the formula.
Now let’s see how we can find K_{b} from K_{a} through the different examples given below.
Solved examples for finding K_{b }from K_{a}
Example #1: The acid dissociation constant (K_{a}) for acetic acid (CH_{3}COOH) at room temperature is 1.8 x 10^{5}. What is the base dissociation constant (K_{b}) for sodium acetate (CH_{3}COO^{–}Na^{+})? 
As per the above statement, K_{a} = 1.8 x 10^{5}. We already know the value of K_{w} at r.t.p, i.e., K_{w} = 1.0 x 10^{14}. We can calculate the value of K_{b} as follows: ⇒ K_{b} = K_{w}/k_{a} ∴ K_{b }= \frac{1.0\times10^{14}}{1.8\times10^{5}} = 5.6 x 10^{10}. Result: The base dissociation constant (K_{b}) for CH_{3}COO^{–}Na^{+} is 5.6 x 10^{10}. Acetic acid and acetate ions are known as conjugate acidbase pairs. 
Example #2: Ammonium (NH_{4}^{+}) ion is a mildly strong conjugate acid of a weak base, i.e., ammonia (NH_{3}). The K_{a }of NH_{4}^{+} is 5.6 x 10^{10}. What is K_{b} for NH_{3}? 
As the K_{a} value is given so we can easily determine the K_{b} value for NH_{3} using the expression K_{b} = K_{w}/K_{a}. ⇒ K_{b} = K_{w}/K_{a} ∴ K_{b} = \frac{1.0\times10^{14}}{5.6\times10^{10}} = 1.8 x 10^{5}. Result: The base dissociation constant (K_{b}) for NH_{3,} as per the above information, is 1.8 x 10^{5}. 
Example # 3: As per Peter’s experimental data, the pH of a 0.500 M formic acid (HCOOH) solution is measured to be 2.04. How can we determine the K_{b} value for formate ion (HCOO^{–}) using this data? 
Formic acid and formate ion are conjugate acidbase pairs. Formic acid dissociates in an aqueous solution, as shown below. So if we first calculate the acid dissociation constant (K_{a}) value for formic acid, we can easily determine the K_{b} value for formate ion. The K_{a} value of formic acid can be calculated using equation (ii), which we learned at the beginning of the article. K_{a} = \frac{[H^{+}][A^{}]}{[HA]} For formic acid, the above equation can be transformed into: K_{a} = \frac{[H^{+}][HCOO^{}]}{[HCOOH]} [H^{+}] can be determined from the pH given in the question statement. pH = log [H^{+}] Making [H^{+}] the subject of the formula. [H^{+}] = 10^{pH} [H^{+}] = 10^{2.04} = 9.12 x 10^{3} M As per the above reaction equation, [H^{+}] = [HCOO^{–}] = 9.12 X 10^{3} M [HCOOH] = formic acid concentration at equilibrium = initial acid concentration – [H^{+}] The initial formic acid concentration is given in the question statement, i.e., 0.500 M. [HCOOH]_{equilibrium }= 0.500 – (9.12 X 10^{3}) = 0.491 M. Finally, we need to plug in all the values determined above into the K_{a} formula. K_{a }= \frac{(9.12\times10^{3})(9.12\times10^{3})}{(0.491)} = 1.7 x 10^{4} Now that we have the K_{a} of formic acid, we can easily determine the K_{b} value for formate ion using K_{b} = K_{w}/K_{a}. ∴ K_{b }= \frac{(1.0\times10^{14})}{(1.7\times10^{4})} = 5.9 × 10^{11}. Result: The base dissociation constant (K_{b}) value for formate ion as per the above data is 5.9 x 10^{11}. 
Example # 4: The pK_{a} for hydrofluoric acid (HF) at 25°C is 3.36. How can we use this information to determine K_{b} for F^{–} (the conjugate base of HF)? 
The pK_{a} value for the acidic solution is related to its K_{a} by the formula given below. pK_{a} = log K_{a} or K_{a} = 10^{pKa} As given in the question statement, pK_{a }for HF = 3.36 so, its K_{a} = 10^{3.36} = 4.4 x 10^{4}. Now that we know K_{a} for HF, we can easily find K_{b} for F^{–} as follows: ∴ K_{b} = K_{w}/k_{a} ∴ K_{b} = \frac{(1.0\times10^{14})}{(4.4\times10^{4})} = 2.2 x 10^{11}. Result: The base dissociation constant (K_{b}) value for fluoride (F^{–}) ion is 2.2 x 10^{11} at 25°C. 
FAQ
What is K_{a}? 
K_{a} is defined as the acid dissociation constant. It determines the extent of ionization of usually a weak acid in an aqueous solution. A higher K_{a} value denotes that the respective acid breaks down to a large extent in water, i.e., it has a higher strength. 
What is K_{b}? 
K_{b }stands for the base dissociation constant that determines the extent of ionization of a base in an aqueous solution. 
How do you differentiate between an acid and a base? 
Acids break down to release H^{+} ions in water, while a base breaks down to produce OH^{–} ions in water. Acids are also defined as proton donors, while bases are proton acceptors as per the Bronsted Lowry acidbase theory.

What is the relationship between K_{a }and K_{b}? 
The product of K_{a} and K_{b} is equal to K_{w,} i.e., the water dissociation constant. K_{w} = K_{a}. K_{b} 
How to find K_{b} from K_{a}? 
The value of K_{w} is fixed at room temperature (25°C), i.e., K_{w} = 1.0 x 10^{14}. If the value of K_{a} of acid is known, then the value for its conjugate base can be determined by rearranging the expression K_{w} = K_{a}. K_{b}, making K_{b} the subject of the formula, as shown below. ∴ K_{b }= K_{w}/K_{a} 
What is a conjugate acidbase pair? 
According to the BronstedLowry theory of acids and bases, a weak acid, such as acetic acid (CH_{3}COOH), dissociates to a small extent in an aqueous solution by releasing H^{+} ions. The acid molecule is converted into an ion by the loss of a proton. This ion is known as the conjugate base of the acid. For e.g. CH_{3}COOH dissociates into acetate (CH_{3}COO^{–}) by losing an H^{+} ion. Consequently, CH_{3}COOH and CH_{3}COO^{–} is known as a conjugate acidbase pair. 
How are K_{a} and K_{b} related to the strength of an acid or a base? 
The higher the K_{a} or K_{b} value, the greater the strength of the respective acid or base. 
Is there a numerical relation between the pH of an acidic solution and its K_{a} value? 
Yes. The K_{a} of an acidic solution can be used to find pK_{a} as follows: ⇒ pK_{a} = log K_{a} pK_{a }is directly related to pH as follows: ⇒ pH = pK_{a} + log [A^{–}]/[HA] In short, the pH of an acidic solution is inversely related to its K_{a} value. The greater the acidic strength, the higher the K_{a} value, but the lower its pH. 
Summary
 The extent of ionization of a weak acid or base in an aqueous solution can be determined using their K_{a} and K_{b} values, respectively.
 K_{a} represents the acid dissociation constant.
 K_{b} stands for base dissociation constant.
 Greater the strength of an acid or base, the higher the respective dissociation constant values because the latter ionizes to a larger extent in an aqueous solution.
 K_{a} is related to K_{b} by the equation K_{w} = K_{a}.K_{b}.
 If the value of K_{a} for acid is known, the K_{b} value for its conjugate base can be determined by rearranging the expression: K_{b} = K_{w}/K_{a}.
 K_{w} = 1.0 x 10^{14} at 25°
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