Answer
a. Weak
$$K_a = \frac{[F^-][H_3O^+]}{[HF]}$$
b. Weak
$$K_a = \frac{[CH{O_2}^-][H_3O^+]}{[HCHO_2]}$$
c. Strong
d, Weak
$$K_a = \frac{[HC{O_3}^-][H_3O^+]}{[H_2CO_3]}$$
Work Step by Step
If the acid is on the "Strong Acids" table (Table 16.3, page 729), it is a strong acid; if the acid does not appear on this list, it is probably a weak acid.
1. Write the balanced equation for the ionization of this acid:
a.$$HNO_3(aq) + H_2O(l) \leftrightharpoons N{O_3}^-(aq) + H_3O^+(aq)$$
b. $$HCHO_2(aq) + H_2O(l) \leftrightharpoons CH{O_2}^-(aq) + H_3O^+(aq)$$
d.$$H_2CO_3(aq) + H_2O(l) \leftrightharpoons HC{O_3}^-(aq) + H_3O^+(aq)$$
2. The exponent of each concentration is equal to its balance coefficient.
a. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ NO{_3}^- ][ H_3O^+ ]}{[ HNO_3 ]}$$
b. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CHO{_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$
d. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ H{CO_3}^- ][ H_3O^+ ]}{[ H_2CO_3 ]}$$
