Answer
$[H^+] = 0.087M$
$pH = 1.06$
Work Step by Step
1000ml = 1L
50.0ml = 0.0500 L
150.0ml = 0.1500 L
1. Find the numbers of moles:
$M(HBr) * V(HBr) = 0.050*0.0500 = 2.5 \times 10^{-3}$ moles
$M(HI) * V(HI) = 0.10*0.1500 = 0.015$ moles
2. Calculate the total volume:
- Total volume: 0.0500 L + 0.1500 L = 0.2000 L
3. Calculate the final concentrations:
$[HBr] = \frac{2.5 \times 10^{-3} mol}{0.2000L} = 0.012M $
$[HI] = \frac{0.015mol}{0.2000L} = 0.075M $
4. Since both acids are strong:
$[H^+]$ produced by $[HBr]$ = $0.012M$
$[H^+]$ produced by $[HI]$ = $0.075M$
Total $[H^+]$= $0.012M + 0.075M = 0.087M$
5. Calculate the pH Value
$pH = -log[H^+]$
$pH = -log( 0.087)$
$pH = 1.06$
