## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 771: 18

#### Answer

When calculating $[H_3O^+]$ for weak acid solutions, we use the $K_a$ expression: $$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$ Substituting for the equilibrium concentrations (we can use an ICE table): $$K_a = \frac{x^2}{[HA]_{initial} - x}$$ In cases where the ionization is not very significant ( less than 5% ), when the acid is very weak or the initial acid concentration is large, we can assume that $$[HA]_{initial} \gt \gt x$$ Meaning that $[HA]_{initial} - x \approx [HA]_{initial}$, which is true considering x is small compared to the acid concentration and the use of significant figures. If we did not use that 'x is small approximation', the expression turns into a quadratic equation; if we use the assumption, the equation is simple and can easily be solved for x.

#### Work Step by Step

When calculating $[H_3O^+]$ for weak acid solutions, we use the $K_a$ expression: $$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$ Substituting for the equilibrium concentrations (we can use an ICE table): $$K_a = \frac{x^2}{[HA]_{initial} - x}$$ In cases where the ionization is not very significant ( less than 5% ), when the acid is very weak or the initial acid concentration is large, we can assume that $$[HA]_{initial} \gt \gt x$$ Meaning that $[HA]_{initial} - x \approx [HA]_{initial}$, which is true considering x is small compared to the acid concentration and the use of significant figures. If we did not use that 'x is small approximation', the expression turns into a quadratic equation; if we use the assumption, the equation is simple and can easily be solved for x.

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