Answer
$pH \approx 12.736$
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:
$n(moles) = V(L) * C = 0.035 * 0.2 = 0.007 moles$
2. Since HBr is a strong acid, and NaOH is a strong base:
$n(HBr) = n(H^+) = 0.004 moles$
$n(NaOH) = n(OH^-) = 0.007 moles$
3. Find the excess of OH-:
*The H+ reacted with OH- in a proportion of 1 to 1.
$0.007 - 0.004 = 0.003 (OH^-)$
4. Convert that number into Concentration:
*Total Volume = 35ml + 20ml = 55ml
$C = \frac{n(moles)}{V(L)} = \frac{0.003}{0.055} \approx 0.0545$
5. Calculate the pOH:
$pOH = -log(0.0545) \approx 1.264$
6. Find the pH:
$pH = 14 - pOH \approx 14 - 1.264 \approx 12.736$
