Chemistry: The Central Science (13th Edition)

$pH = 10.698$
1. Find the number of moles of HBr and NaOH: HBr: $n(moles) = V(L) * C = 0.02 * 0.2 = 0.004 moles$ NaOH: $n(moles) = V(L) * C = 0.0201 * 0.2 = 0.00402 moles$ 2. Since HBr is a strong acid, and NaOH is a strong base: $n(HBr) = n(H^+) = 0.004 moles$ $n(NaOH) = n(OH^-) = 0.00402 moles$ 3. Find the excess of OH-: *The H+ reacted with OH- in a proportion of 1 to 1. $0.00402 - 0.004 = 0.00002 (OH^-)$ 4. Convert that number into Concentration: *Total Volume = 20.1ml + 20ml = 40.1ml $C = \frac{n(moles)}{V(L)} = \frac{0.00002}{0.0401} = 4.988 \times 10^{-4}M$ 5. Calculate the pOH: $pOH = -log(4.988 \times 10^{-4}) \approx 3.302$ 6. Find the pH: $pH = 14 - pOH = 14 - 3.302 = 10.698$