Answer
At that point, the pH of the solution is equal to 2.07
Work Step by Step
- As we have calculated in the exercise 69b, the volume of added $HCl$ at the equivalence point is equal to 28.8 mL. If we add 5.0 mL to that volume, we get $33.8$ $mL$ of that base solution.
1000ml = 1L
33.8ml = 0.0338 L
25.0ml = 0.0250 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.100* 0.0338 = 3.38 \times 10^{-3}$ moles
$C(RbOH) * V(RbOH) = 0.115* 0.0250 = 2.88 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + RbOH(aq) -- \gt RbCl(aq) + H_2O(l)$
- Total volume: 0.0338 + 0.0250 = 0.0588L
3. Since the base is the limiting reactant, only $ 0.002875$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.00338 - 0.002875 = 5.0 \times 10^{-4}$ moles.
Concentration: $\frac{5.0 \times 10^{-4}}{ 0.0588} = 8.5 \times 10^{-3}M$
$[RbOH] = 0.002875 - 0.002875 = 0 $ moles
- Since $HCl$ is a strong acid:
$[HCl] = [H_3O^+] = 8.5 \times 10^{-3}M$
4. Calculate the pH value:
$pH = -log[H_3O^+]$
$pH = -log( 8.5 \times 10^{- 3})$
$pH = 2.07$
