Answer
(a) $pH = 11.85$
(b) $pH = 1.426$
Work Step by Step
(a)
1000ml = 1L
15ml = 0.015 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.35* 0.015 = 5.25 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.275* 0.02 = 5.5 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$
- Total volume: 0.015 + 0.02 = 0.035L
3. Since the acid is the limiting reactant, only $ 0.00525$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.00525 - 0.00525 = 0M$.
$[KOH] = 0.0055 - 0.00525 = 2.5 \times 10^{-4}$ mol
Concentration: $\frac{2.5 \times 10^{-4}}{ 0.035} = 7.1 \times 10^{-3}M$
- The only significant electrolyte in the solution is $KOH$, which is a strong base, so:
$[OH^-] = [KOH] = 7.1 \times 10^{-3}M$
4. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 7.1 \times 10^{- 3})$
$pOH = 2.15$
5. Find the pH:
$pH + pOH = 14$
$pH + 2.15 = 14$
$pH = 11.85$
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(b)
1000ml = 1L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.35* 0.02 = 7 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.275* 0.02 = 5.5 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$
- Total volume: 0.02 + 0.02 = 0.04L
3. Since the base is the limiting reactant, only $ 0.0055$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.007 - 0.0055 = 1.5 \times 10^{-3}$ moles.
Concentration: $\frac{1.5 \times 10^{-3}}{ 0.04} = 0.0375M$
$[KOH] = 0.0055 - 0.0055 = 0 $ moles
- Since : $HCl$ is a strong acid:
$[HCl] = [H_3O^+] = 0.0375M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.0375)$
$pH = 1.426$
