Answer
$pH \approx 0.778$
Work Step by Step
1. Find the number of moles of $HBr$ and $HCl$:
$HBr:$
$n(moles) = Concentration \times Volume(L)$
$n(moles) = 0.100 \times 0.01$
$n(moles) = 0.001$
$HCl:$
$n(moles) =0.200 \times 0.02$
$n(moles) = 0.004$
2. Find the number of $H^+$ moles:
Since $HBr$ and $HCl$ are strong acids:
$n(H^+) = n(HBr) + n(HCl)$
$n(H^+) = 0.001 + 0.004$
$n(H^+) = 0.005 moles$
3. Find the concentration of $H^+$
$Total Volume = 20ml + 10ml = 30ml$
$Concentration = \frac{n(moles)}{V(L)}$
$[H^+] = \frac{0.005}{0.030}$
$[H^+] \approx 0.167$
3. Calculate the pH:
$pH = -log[H^+]$
$pH \approx -log(0.167)$
$pH \approx 0.778$
