Given that Ka for HClO is 4.0 × 10-8 at 25 °C, what is the value of Kb for ClO– at 25 °C?
Given that Kb for CH3CH2NH2 is 6.3 × 10-4 at 25 °C, what is the value of Ka for CH3CH2NH3 at 25 °C?
KaKb = Kw
To find the value of Kb for ClO- at 25 °C, we can use the relationship between Ka and Kb for the conjugate acid-base pair.
The relationship between Ka and Kb is given by the equation: Ka x Kb = Kw
Where Kw is the ionization constant of water at a given temperature. At 25 °C, the value of Kw is 1.0 x 10^-14.
Since we know the value of Ka for HClO, we can substitute it into the equation:
Ka x Kb = Kw
(4.0 × 10^-8) x Kb = 1.0 x 10^-14
Now, let's solve for Kb:
Kb = Kw / Ka
Kb = (1.0 x 10^-14) / (4.0 × 10^-8)
Kb = 2.5 x 10^-7
Therefore, the value of Kb for ClO- at 25 °C is 2.5 x 10^-7.
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To find the value of Ka for CH3CH2NH3 at 25 °C, we will once again use the relationship between Ka and Kb.
Using the equation Ka x Kb = Kw, we can rearrange to solve for Ka:
Ka = Kw / Kb
We are given the value of Kb for CH3CH2NH2, so we can substitute it into the equation:
Ka = Kw / Kb
Ka = (1.0 x 10^-14) / (6.3 × 10^-4)
Now, let's solve for Ka:
Ka = 1.6 x 10^-11
Therefore, the value of Ka for CH3CH2NH3 at 25 °C is 1.6 x 10^-11.