Answer
$pH = 10.698$
Work Step by Step
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:
$pH = 14 - pOH = 14 - 3.302 = 10.698$
