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Determine the ka of the weak acid ha whose conjugate base has kb 2. 9 × 10−8

Sagot :

Vmonicaav80idn

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! ♡

Mahima6824soniidn

The ka of the weak acid ha, whose conjugate base has kb 2. 9 × 10⁻⁸ is 3.45 × 10⁻⁷.

What is ka value?

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.

Learn more about ka values

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