Answer
a. $pH = 0.699$
b. $pH = 0.854$
c. $pH = 1.301$
d. $pH = 7.00$
e. $pH = 12.155$
Work Step by Step
(a)
- Since : $HClO_4$ is a strong acid, and we have a pure solution:
$[HClO_4] = [H_3O^+] = 0.2M$
1. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.2)$
$pH = 0.699$
(b)
1000ml = 1L
40ml = 0.04 L
10ml = 0.01 L
1. Find the numbers of moles:
$C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.01 = 1 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$
- Total volume: 0.04 + 0.01 = 0.05L
3. Since the base is the limiting reactant, only $ 0.001$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HClO_4] = 0.008 - 0.001 = 7 \times 10^{-3}$ moles.
Concentration: $\frac{7 \times 10^{-3}}{ 0.05} = 0.14M$
$[KOH] = 0.001 - 0.001 = 0 $ moles
- Since : $HClO_4$ is a strong acid:
$[HClO_4] = [H_3O^+] = 0.14M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.14)$
$pH = 0.8539$
(c)
1000ml = 1L
40ml = 0.04 L
40ml = 0.04 L
1. Find the numbers of moles:
$C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.04 = 4 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$
- Total volume: 0.04 + 0.04 = 0.08L
3. Since the base is the limiting reactant, only $ 0.004$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HClO_4] = 0.008 - 0.004 = 4 \times 10^{-3}$ moles.
Concentration: $\frac{4 \times 10^{-3}}{ 0.08} = 0.05M$
$[KOH] = 0.004 - 0.004 = 0 $ moles
- Since : $HClO_4$ is a strong acid:
$[HClO_4] = [H_3O^+] = 0.05M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.05)$
$pH = 1.301$
(d)
1000ml = 1L
40ml = 0.04 L
80ml = 0.08 L
1. Find the numbers of moles:
$C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.08 = 8 \times 10^{-3}$ moles
2. When the number of moles is equal, the reactants are totally consumed:
$HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$
- Total volume: 0.04 + 0.08 = 0.12L
3. So, those are the final concentrations:
$[HClO_4] = 0.008 - 0.008 = 0$ mol.
$[KOH] = 0.008 - 0.008 = 0$ mol
- Therefore, the solution doesn't have any significant electrolyte: pH = 7 (Neutral)
(e)
1000ml = 1L
40ml = 0.04 L
100ml = 0.1 L
1. Find the numbers of moles:
$C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.1* 0.1 = 0.01$ moles
2. Write the acid-base reaction:
$HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$
- Total volume: 0.04 + 0.1 = 0.14L
3. Since the acid is the limiting reactant, only $ 0.008$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HClO_4] = 0.008 - 0.008 = 0M$.
$[KOH] = 0.01 - 0.008 = 2 \times 10^{-3}$ mol
Concentration: $\frac{2 \times 10^{-3}}{ 0.14} = 0.01429M$
- The only significant electrolyte in the solution is $KOH$, which is a strong base, so:
$[OH^-] = [KOH] = 0.01429M$
$pOH = -log[OH^-]$
$pOH = -log( 0.01429)$
$pOH = 1.845$
$pH + pOH = 14$
$pH + 1.845 = 14$
$pH = 12.155$
