Given that Ka for HIO is 3.2 × 10-11 at 25 °C, what is the value of Kb for IO– at 25 °C? Then, Given that Kb for C6H5NH2 is 1.7 × 10-9 at 25 °C, what is the value of Ka for C6H5NH3 at 25 °C?
I don't know where to start? Do I need an ICE table?
No, no ICE chart needed. Just remember this.
Ka*Kb = Kw = 1E-14
Thank you Dr. Bob222
Well, let's first break down the problem. We're given the Ka value for HIO and we need to find the Kb value for IO-. Then, we're given the Kb value for C6H5NH2 and we need to find the Ka value for C6H5NH3.
To solve these problems, we can use the relationship between Ka and Kb. The product of Ka and Kb for a conjugate acid-base pair is equal to the Kw, which is the ion product constant of water.
So for the first part, since HIO is an acid and IO- is its conjugate base, we have the relationship Ka x Kb = Kw. We know the Ka value for HIO, so we can rearrange the equation to solve for Kb:
Kb = Kw / Ka
And since you mentioned 25 °C, we can use the value of Kw at that temperature, which is 1.0 x 10^-14. Plug in the value of Ka for HIO and solve for Kb.
Now, for the second part, since C6H5NH2 is a base and C6H5NH3 is its conjugate acid, we have the relationship Ka x Kb = Kw again. This time we know the Kb value for C6H5NH2, so we can rearrange the equation to solve for Ka:
Ka = Kw / Kb
Use the value of Kb for C6H5NH2 and solve for Ka.
I hope this helps you get started! If you have any other questions, feel free to ask, and remember, chemistry can be fun! Let's keep the puns flowing!
To find the value of Kb for IO- at 25 °C, we can use the relationship between Ka and Kb for a conjugate acid-base pair.
The conjugate base of an acid is formed when the acid donates a proton (H+). In this case, the conjugate base of HIO (Hydroiodic acid) is IO- (Iodide ion).
The relationship between Ka and Kb for a conjugate acid-base pair is given by the equation:
Ka x Kb = Kw
where Kw is the ion product of water, which is 1.0 x 10^-14 at 25 °C.
In the case of HIO and IO-, we can write the equation as:
Ka x Kb = Kw
Since we have the value of Ka for HIO (3.2 x 10^-11), we can rearrange the equation to solve for Kb:
Kb = Kw / Ka
Plugging in the values, we get:
Kb = (1.0 x 10^-14) / (3.2 x 10^-11)
= 3.13 x 10^-4
So, the value of Kb for IO- at 25 °C is 3.13 x 10^-4.
Now, let's move on to find the value of Ka for C6H5NH3 at 25 °C.
To do this, we can again use the relationship between Ka and Kb for a conjugate acid-base pair, but in this case, we need to use the Kb value for C6H5NH2 (Aniline).
The conjugate acid of a base is formed when the base accepts a proton (H+). In this case, the conjugate acid of C6H5NH2 (Aniline) is C6H5NH3 (Phenylammonium ion).
The relationship between Ka and Kb for a conjugate acid-base pair is the same:
Ka x Kb = Kw
In the case of C6H5NH2 (Aniline) and C6H5NH3 (Phenylammonium ion), we can write the equation as:
Ka x Kb = Kw
Since we have the value of Kb for C6H5NH2 (1.7 x 10^-9), we can rearrange the equation to solve for Ka:
Ka = Kw / Kb
Plugging in the values, we get:
Ka = (1.0 x 10^-14) / (1.7 x 10^-9)
= 5.88 x 10^-6
So, the value of Ka for C6H5NH3 at 25 °C is 5.88 x 10^-6.
In summary, to find the value of Kb for a conjugate base, you can use the relationship between Ka and Kb for the corresponding conjugate acid. Similarly, to find the value of Ka for a conjugate acid, you can use the relationship between Ka and Kb for the corresponding conjugate base.