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Answer: [tex]3.45*10^{-7}[/tex]
Explanation:
we can use the equation Ka * Kb = Kw
Kw is the equilibrium constant and always equals to [tex]1*10^{-14}[/tex]
we plug the given numbers into the equation, [tex]Ka * (2.9*10^{-8} )=(1*10^{-14} )[/tex]
now we just solve for Ka, [tex]Ka=\frac{1*10^{-14} }{2.9*10^{-8} }[/tex]
which gives us the value, [tex]Ka=3.45*10^{-7}[/tex]
hope this helped! ♡
The ka of the weak acid ha, whose conjugate base has kb 2. 9 × 10⁻⁸ is 3.45 × 10⁻⁷.
Ka value is the equilibrium constant found by the dissociation of acid.
By the equation
[tex]\rm ka \times kb = kw[/tex]
where kb = [tex]2.9 \times 10^-^8[/tex]
kw = [tex]1 \times 10^-^4[/tex]
Now, putting the values in the equation
[tex]1 \times 10^-^1^4 = ka \times 2.9 \times 10^-^8\\\\ka = \dfrac{1 \times 10^-^1^4}{2.9 \times 10^-^8} \\\\ka = 3.45 \times 10^-^7[/tex]
Thus, the ka of the weak acid ha, whose conjugate base has kb 2. 9 × 10⁻⁸ is 3.45 × 10⁻⁷.
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