Answer
$pH = 10.70$
Work Step by Step
1000ml = 1L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.1* 0.05 = 5 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.1* 0.0505 = 5.2 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$
- Total volume: 0.05 + 0.0505 = 0.1005L
3. Since the acid is the limiting reactant, only $ 0.005$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.005 - 0.005 = 0M$.
$[NaOH] = 0.00505 - 0.005 = 5 \times 10^{-5}$ mol
Concentration: $\frac{5.0 \times 10^{-5}}{ 0.1005} = 5 \times 10^{-4}M$
- The only significant electrolyte in the solution is $NaOH$, which is a strong base, so:
$[OH^-] = [NaOH] = 5.0 \times 10^{-4}M$
4. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 5.0 \times 10^{- 4})$
$pOH = 3.30$
5. Find the pH:
$pH + pOH = 14$
$pH + 3.3 = 14$
$pH = 10.70$
