Consider the following equilibrium:
$$
\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \Leftrightarrow \mathrm{HC}_{2} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{~S} \quad \mathrm{~K}_{\mathrm{eq}}=3.1 \times 10^{1}
$$

The stronger acid is
$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
$\mathrm{H}_{2} \mathrm{~S}$
$\mathrm{HC}_{2} \mathrm{O}_{4}^{-}$
$\mathrm{HS}^{-}$

Question

Consider the following equilibrium:
$$
\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \Leftrightarrow \mathrm{HC}_{2} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{~S} \quad \mathrm{~K}_{\mathrm{eq}}=3.1 \times 10^{1}
$$

The stronger acid is
$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$
$\mathrm{H}_{2} \mathrm{~S}$
$\mathrm{HC}_{2} \mathrm{O}_{4}^{-}$
$\mathrm{HS}^{-}$

Accepted Answer

***B*** ∻Key Concept∻ ⚹Acid strength and equilibrium position⚹ ∻Explanation∻ ⚹The stronger acid will donate a proton more readily, shifting the equilibrium towards the products. Given the equilibrium constant (Keq) is greater than 1, the reaction favors the formation of products, indicating that $\mathrm{H_2C_2O_4}$ is the stronger acid compared to $\mathrm{HS^-}$.⚹◊What is the relationship between acid strength and the equilibrium constant in a chemical reaction?⍭ Generate me a similar question◊

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