Answer
$C, F$ and $G$.
Work Step by Step
1. Since we are comparing the pH value, identify all the solutions in the grid:
$C : [OH^-] = 3.0 \times 10^{-5}M$
$F : 0.10M LiOH$
$G: [H_3O^+] = 2.00 \times 10^{-8}M$
$I: 0.010M HClO_4$
2. Now, calculate the pH for each one:
$C$:
$pOH = -log[OH^-]$
$pOH = -log( 3 \times 10^{- 5})$
$pOH = 4.523$
$pH + pOH = 14$
$pH + 4.523 = 14$
$pH = 9.477$
$9.477 > 7$; Therefore, this is one of the solutions.
$F:$
- $LiOH$ is a strong base, so: $[OH^-] = [LiOH] = 0.10M$
$pOH = -log[OH^-]$
$pOH = -log( 0.1)$
$pOH = 1$
$pH + pOH = 14$
$pH + 1 = 14$
$pH = 13$
$13 > 7$; Therefore this is one of the solutions
$G:$
$pH = -log[H_3O^+]$
$pH = -log( 2 \times 10^{- 8})$
$pH = 7.699$
$7.699 > 7$; Therefore, this is one of the solutions.
$I:$
$HClO_4$ is a strong acid, so: $[H_3O^+] = [HClO_4] = 0.010M$
$pH = -log[H_3O^+]$
$pH = -log( 0.01)$
$pH = 2$
$2 < 7$; Not valid.
