Chemistry: Molecular Approach (4th Edition)

a. $$pH = -log[H_3O^+] = -log( 9.5 \times 10^{-9} ) = 8.02$$ Since pH $\gt$ 7, the solution is basic. b. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 7.1 \times 10^{-3} } = 1.4 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 1.4 \times 10^{-12} ) = 11.85$$ Since pH $\gt$ 7, the solution is basic.