Answer
The pH of the solution after 10.0 mL of added base is equal to 1.038
Work Step by Step
1000ml = 1L
35.0ml = 0.0350 L
10.0ml = 0.0100 L
1. Find the numbers of moles:
$C(HBr) * V(HBr) = 0.175* 0.035 = 6.13 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.200* 0.0100 = 2.00 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HBr(aq) + KOH(aq) -- \gt KBr(aq) + H_2O(l)$
- Total volume: 0.0350 + 0.0100 = 0.0450L
3. Since the base is the limiting reactant, only $ 0.002$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HBr] = 0.006125 - 0.00200 = 4.125 \times 10^{-3}$ moles.
Concentration: $\frac{4.125 \times 10^{-3}}{ 0.0450} = 0.0917M$
$[KOH] = 0.00200 - 0.00200 = 0 $ moles
- Since : $HBr$ is a strong acid:
$[HBr] = [H_3O^+] = 0.0917M$
4. Calculate the pH value
$pH = -log[H_3O^+]$
$pH = -log( 0.0917)$
$pH = 1.038$
