Answer
The pH of the solution at the point is equal to 12.90
Work Step by Step
1000ml = 1L
5.0ml = 0.0050 L
25.0ml = 0.0250 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.100* 0.0050 = 5.0 \times 10^{-4}$ 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: 0.0050 + 0.0250 = 0.0300L
3. Since the acid is the limiting reactant, only $ 5 \times 10^{-4}$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.0005 - 0.0005 = 0M$.
$[RbOH] = 0.002875 - 0.0005 = 2.375 \times 10^{-3}$ mol
Concentration: $\frac{2.375 \times 10^{-3}}{ 0.0300} = 0.0792M$
- The only significant non-neutral compound in the solution is $RbOH$, which is a strong base, so:
$[OH^-] = [RbOH] = 0.0792M$
4. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 0.0792)$
$pOH = 1.10$
5. Find the pH:
$pH + pOH = 14$
$pH + 1.10 = 14$
$pH = 12.90$
