Answer
(a) $pH = 0.996$
(b) $pH = 1.602$
(c) $pH = 12.57$
Work Step by Step
(a)
1000ml = 1L
25ml = 0.025 L
3ml = $3 \times 10^{-3} L
$
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.125* 0.025 = 3.13 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.003 = 3 \times 10^{-4}$ moles
2. Write the acid-base reaction:
$HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$
- Total volume: 0.025 + 0.003 = 0.028L
3. Since the base is the limiting reactant, only $ 0.0003$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.003125 - 0.0003 = 2.83 \times 10^{-3}$ moles.
Concentration: $\frac{2.83 \times 10^{-3}}{ 0.028} = 0.101M$
$[KOH] = 0.0003 - 0.0003 = 0 $ moles
- Since $HCl$ is a strong acid:
$[HCl] = [H_3O^+] = 0.101M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.101)$
$pH = 0.9961$
(b)
1000ml = 1L
25ml = 0.025 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.125* 0.025 = 3.13 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.02 = 2 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$
- Total volume: 0.025 + 0.02 = 0.045L
3. Since the base is the limiting reactant, only $ 0.002$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.003125 - 0.002 = 1.12 \times 10^{-3}$ moles.
Concentration: $\frac{1.12 \times 10^{-3}}{ 0.045} = 0.025M$
$[KOH] = 0.002 - 0.002 = 0 $ moles
- Since $HCl$ is a strong acid:
$[HCl] = [H_3O^+] = 0.025M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.025)$
$pH = 1.602$
(c)
1000ml = 1L
25ml = 0.025 L
65ml = 0.065 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.125* 0.025 = 3.13 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.065 = 6.5 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$
- Total volume: 0.025 + 0.065 = 0.09L
3. Since the acid is the limiting reactant, only $ 0.003125$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.003125 - 0.003125 = 0M$.
$[KOH] = 0.0065 - 0.003125 = 3.38 \times 10^{-3}$ mol
Concentration: $\frac{3.38 \times 10^{-3}}{ 0.09} = 0.0375M$
- The only significant electrolyte in the solution is $KOH$, which is a strong base, so:
$[OH^-] = [KOH] = 0.0375M$
$pOH = -log[OH^-]$
$pOH = -log( 0.0375)$
$pOH = 1.43$
$pH + pOH = 14$
$pH + 1.43 = 14$
$pH = 12.57$
