Question

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

Answer 1

Answer:  $3.45*10^{-7}$

Explanation:

we can use the equation Ka * Kb = Kw

Kw is the equilibrium constant and always equals to $1*10^{-14}$

we plug the given numbers into the equation, $Ka * (2.9*10^{-8} )=(1*10^{-14} )$

now we just solve for Ka, $Ka=\frac{1*10^{-14} }{2.9*10^{-8} }$

which gives us the value, $Ka=3.45*10^{-7}$

hope this helped! ♡

Answer 2

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

$\rm ka \times kb = kw$

where kb = $2.9 \times 10^-^8$

kw = $1 \times 10^-^4$

Now, putting the values in the equation

$1 \times 10^-^1^4 = ka \times 2.9 \times 10^-^8\\\\ka = \dfrac{1 \times 10^-^1^4}{2.9 \times 10^-^8} \\\\ka = 3.45 \times 10^-^7$

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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