Answer
$[OH^-] \approx 8.75 \times 10^{-5}M$
$pH \approx 9.942$
Work Step by Step
1. Find $[NaOH]$ after the dilution:
$C_i * V_i = C_f * V_f$
$0.175 * 0.001 = C_f * 2$
$C_f = 8.75 \times 10^{-5}$
2. Since $NaOH$ is a strong base:
$[NaOH] = [OH^-] = 8.75 \times 10^{-5}$
3. Find the pH:
$pOH = -log[OH^-]$
$pOH = -log(8.75 \times 10^{-5})$
$pOH = 4.058$
$pH = 14 - pOH$
$pH = 14 - 4.058 = 9.942$
