The value of Kb for HPO₄²- is?
A. 2.2*10^-13
B. 6.2*10^-8
C. 1.6*10^-7
D. 4.5*10^-2
I don't get how to solve this.
With that being said,
Kb=Kw/Ka= 1 x 10^-14/6.2 x10^-8 =1.6 x10^-7
How do you know which ka to use? There is one on the base side of the chart, and one of the acid side of the chart. For the one of the acid side of the chart, the Ka is 2.2*10^-13 And for the one on the base side of the chart, it is 6.2*10^-8.
Correct me if I'm wrong but
HPO4^2- + HOH ==> H2PO4^- + OH^-
k3 for H3PO4= 4.2E-13 = (H^+)(PO4^3-)/(HPO4^-) and that doesn't contain the ions needed for HPO4^2- to act as a base. k2 for H3PO4 does.
k2 = (H^+)(HPO4^2-)/(H2PO4^-)
I think the answer is c.
Its c, I used the wrong Ka value when I initially did it.
Thanks so much :) So if we were to find the Ka, then we would use the ka on the acid side of the chart?
Yes, the way your reference is set-up. Remember, Kw= Ka*Kb. if you want to know the Kb, you need to know the Kw and Ka. If you want to know the Kb, you need to know the Ka and Kw.
Thanks soo much to both of you :)
To find the value of Kb for HPO₄²-, we can use the concept of chemical equilibrium. Kb is the equilibrium constant for the reaction when the ion HPO₄²- acts as a base and accepts a proton, forming the conjugate acid.
The chemical equation for this equilibrium can be written as:
HPO₄²- + H₂O ⇌ H₂PO₄- + OH-
Since the reaction is happening in water, we assume that water is the solvent and it remains in its pure liquid form. So, we can ignore the concentration of water from the equilibrium expression.
Thus, the Kb expression for HPO₄²- can be written as:
Kb = [H₂PO₄-][OH-] / [HPO₄²-]
To solve this problem, we need to know the concentration of H₂PO₄-, OH-, and HPO₄²-. Unfortunately, the concentrations are not given in the question.
However, we can still solve this problem by using the relationship between Kb and Ka (the equilibrium constant for the conjugate acid). The value of Ka for H₂PO₄- is known to be 6.3 x 10^-8.
The relationship between Kb and Ka is given by the expression:
Kw = Ka * Kb
W is the equilibrium constant for the self-ionization of water and is equal to 1.0 x 10^-14 at 25°C.
Substituting the known values, we have:
1.0 x 10^-14 = (6.3 x 10^-8) * Kb
Now, rearrange the equation to solve for Kb:
Kb = (1.0 x 10^-14) / (6.3 x 10^-8)
By plugging these values into a calculator, we get:
Kb ≈ 1.6 x 10^-7
Hence, the value of Kb for HPO₄²- is approximately 1.6 x 10^-7.
Therefore, the correct answer is option C: 1.6 x 10^-7.
It's Kw/k2 for H3PO4.
I'm sure you remember Kw = 1E-14.
If you are asking what is the dissociation constant (Kb) for HPO₄², you want be able to solve for it with the information given; you have to look up the Ka for HPO₄² in a book.
Ka for HPO₄²=4.2 x10^-13
Kw= Ka*Kb
Solving for Kb,
Kb=Kw/Ka= 1 x 10^-14/4.2 x10^-13 =2.40 x 10^-2
This answer is not one of your answer choices, so I'm a little stuck on this one.