Answer
The pH at that point is equal to 12.15
Work Step by Step
- As we have calculated in the exercise 67b, the volume of added $KOH$ at the equivalence point is equal to 30.6 mL. If we add 5.0 mL to that volume, we got $35.6$ $mL$ of that base solution.
1000ml = 1L
35.0ml = 0.0350 L
35.6ml = 0.0356 L
1. Find the numbers of moles:
$C(HBr) * V(HBr) = 0.175* 0.0350 = 6.125 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.200* 0.0356 = 7.12 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HBr(aq) + KOH(aq) -- \gt KBr(aq) + H_2O(l)$
- Total volume: 0.0350 + 0.0356 = 0.0706L
3. Since the acid is the limiting reactant, only $ 0.006125$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HBr] = 0.006125 - 0.006125 = 0M$.
$[KOH] = 0.00712 - 0.006125 = 9.95 \times 10^{-4}$ mol
Concentration: $\frac{9.95 \times 10^{-4}}{ 0.0706} = 0.0141M$
- The only significant electrolyte in the solution is $KOH$, which is a strong base, so:
$[OH^-] = [KOH] = 0.0141M$
4. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 0.0141)$
$pOH = 1.85$
5. Find the pH:
$pH + pOH = 14$
$pH + 1.85 = 14$
$pH = 12.15$
