Question

for the following reaction at equilibrium
hclo₄ + h₂o → clo₄⁻¹ + h₃o⁺¹
hclo₄ is the 1 . it is on the list of strong acids and so is a 2 acid.
h₂o is the 3 . it is both a very 4 acid and a very 5 base.
clo₄⁻¹ is the 6 . it does not contain hydroxide so is a 7 base.
h₃o⁺¹ is the 8 . it is a 9 acid because it is on the list of strong acids.
the kₐ approaches positive infinity.
note: we treat moderately strong as weak in this course
a. acid b. base c. conjugate acid d. conjugate base
e. strong f. weak

Explanation:

1. In the reaction \( \ce{HClO4 + H2O -> ClO4^{-} + H3O^{+}} \), \( \ce{HClO4} \) donates a proton (\( \ce{H^{+}} \)) to \( \ce{H2O} \), so it is an acid (A). Since it's a strong acid (given it's on the strong acids list), it's a strong (E) acid.
2. \( \ce{H2O} \) accepts a proton from \( \ce{HClO4} \), so it is a base (B). Water is a very weak (F) acid and a very weak (F) base (in the context of this course, moderately strong is treated as weak, and water is a weak acid and weak base).
3. \( \ce{ClO4^{-}} \) is the conjugate base (D) of \( \ce{HClO4} \) (since \( \ce{HClO4} \) lost a proton to form it). It doesn't have \( \ce{OH^{-}} \) and is a weak (F) base (as it's the conjugate base of a strong acid, which makes it a weak base).
4. \( \ce{H3O^{+}} \) is the conjugate acid (C) of \( \ce{H2O} \) (since \( \ce{H2O} \) gained a proton to form it). It's a strong (E) acid (as it's on the strong acids list).

Answer:

1. A. acid; E. strong
2. B. base; F. weak; F. weak
3. D. conjugate base; F. weak
4. C. conjugate acid; E. strong