Answer
(a) $pH = 12.99$
(b) $pH = 12.31$
(c) $pH = 1.95$
Work Step by Step
(a)
1000ml = 1L
5ml =$ 5 \times 10^{-3} L$
50ml = 0.05 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.075* 0.005 = 3.75 \times 10^{-4}$ moles
$C(NaOH) * V(NaOH) = 0.116* 0.05 = 5.8 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$
- Total volume: 0.005 + 0.05 = 0.055L
3. Since the acid is the limiting reactant, only $ 0.000375$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.000375 - 0.000375 = 0M$.
$[NaOH] = 0.0058 - 0.000375 = 5.43 \times 10^{-3}$ mol
Concentration: $\frac{5.43 \times 10^{-3}}{ 0.055} = 0.0985M$
- The only significant electrolyte in the solution is $NaOH$, which is a strong base, so:
$[OH^-] = [NaOH] = 0.0985M$
$pOH = -log[OH^-]$
$pOH = -log( 0.0986)$
$pOH = 1.01$
$pH + pOH = 14$
$pH + 1.01 = 14$
$pH = 12.99$
(b)
1000ml = 1L
50ml = 0.05 L
50ml = 0.05 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.075* 0.05 = 3.75 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.116* 0.05 = 5.8 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$
- Total volume: 0.05 + 0.05 = 0.1L
3. Since the acid is the limiting reactant, only $ 0.00375$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.00375 - 0.00375 = 0M$.
$[NaOH] = 0.0058 - 0.00375 = 2.04 \times 10^{-3}$ mol
Concentration: $\frac{2.04 \times 10^{-3}}{ 0.1} = 0.0205M$
- The only significant electrolyte in the solution is $NaOH$, which is a strong base, so:
$[OH^-] = [NaOH] = 0.0205M$
$pOH = -log[OH^-]$
$pOH = -log( 0.0205)$
$pOH = 1.69$
$pH + pOH = 14$
$pH + 1.69 = 14$
$pH = 12.31$
(c)
1000ml = 1L
50ml = 0.05 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.075* 0.1 = 7.5 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.116* 0.05 = 5.8 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$
- Total volume: 0.1 + 0.05 = 0.15L
3. Since the base is the limiting reactant, only $ 0.0058$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.0075 - 0.0058 = 1.6 \times 10^{-3}$ moles.
Concentration: $\frac{1.6 \times 10^{-3}}{ 0.15} = 0.0112M$
$[NaOH] = 0.0058 - 0.0058 = 0 $ moles
- Since $HCl$ is a strong acid:
$[HCl] = [H_3O^+] = 0.0112M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.0113)$
$pH = 1.95$
