Answer
$pH \approx 3.3$
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:
$n(moles) = V(L) * C = 0.0199 * 0.2 = 0.00398 moles$
2. Since HBr is a strong acid, and NaOH is a strong base:
$n(HBr) = n(H^+) = 0.004 moles$
$n(NaOH) = n(OH^-) = 0.00398 moles$
3. Find the excess of H+:
*The H+ reacted with OH- in a proportion of 1 to 1.
$0.004 - 0.00398 = 0.00002 (H^+)$
4. Convert that number into Concentration:
*Total Volume = 19.9ml + 20ml = 39.9ml
$C = \frac{n(moles)}{V(L)} = \frac{0.00002}{0.0399} = 0.0005013M$
5. Calculate the pH:
$pH = -log(0.0005013) \approx 3.3$
