At 25^{} C, K_{a} of formic acid is 1.8 × 10^{-4}. What is the K_{b} of {HCOO}^{-}? [AP EAPCET 22 May 2025 S2]

Correct Answer is : 5.55×10−115.55 × 10^{-11}5.55×10−11

Correct Answer is : 5.55×10−115.55 × 10^{-11}5.55×10−11

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Ionic Equilibrium

Section: Chemistry

Question : 3 of 14

Marks: +1, -0

25^{\circ} C, K_{a}

1.8 \times 10^{-4}

K_{b}

\mathrm{HCOO}^{-}

Solution:

The ion product of water

(K_w)

is the product of the acid dissociation constant

(K_a)

and the base dissociation constant

(K_b)

K_a \times K_b = K_w

So, you can solve for

K_b

using:

K_b=\;\frac{K_w}{K_a}

We know:

\;K_w=1.0 \times 10^{-14} \; \text{ at } \; 25^{\circ} C

\;K_a=1.8 \times 10^{-4}

Plug these values into the formula:

\;K_b=\;\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}}

\;K_b=5.55 \times 10^{-11}

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