## Chemistry (4th Edition)

(a) $[LiOH] = 6.02 \times 10^{- 5}M$ (b) $[Ba(OH)_2] = 3.01 \times 10^{-5}M$
1. Find $[OH^-]$ pH + pOH = 14 9.78 + pOH = 14 pOH = 4.22 $[OH^-] = 10^{- 4.22}$ $[OH^-] = 6.02 \times 10^{- 5}M$ ------ (a) Since LiOH is a strong base: $[LiOH] = [OH^-] = 6.02 \times 10^{- 5}M$ (b) Since $Ba(OH)_2$ is a strong base with 2 $OH^-$: $[OH^-] = [Ba(OH)_2] * 2$ $6.02 \times 10^{- 5} = [Ba(OH)_2] * 2$ $\frac{ 6.02 \times 10^{- 5}}{2} = [Ba(OH)_2]$ $[Ba(OH)_2] = 3.01 \times 10^{-5}M$