Answer
(a) $pH = 2.87$
(b) $pH = 3.51$
Work Step by Step
(a)
1000ml = 1L
25ml = 0.025 L
10ml = 0.01 L
1. Find the numbers of moles:
$C(HNO_2) * V(HNO_2) = 0.132* 0.025 = 3.3 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.116* 0.01 = 1.15 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HNO_2(aq) + NaOH(aq) -- \gt NaNO_2(aq) + H_2O(l)$
- Total volume: 0.025 + 0.01 = 0.035L
3. Since the base is the limiting reactant, only $ 0.00116$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HNO_2] = 0.0033 - 0.00116 = 2.14 \times 10^{-3}$ moles.
Concentration: $\frac{2.14 \times 10^{-3}}{ 0.035} = 0.0611M$
$[NaOH] = 0.00116 - 0.00116 = 0$
$[NaNO_2] = 0 + 0.00116 = 0.00116$ moles.
Concentration: $\frac{ 0.00116}{ 0.035} = 0.0332M$
4. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 7.2 \times 10^{- 4})$
$pKa = 3.14$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.14 + log(\frac{0.0332}{0.0611})$
$pH = 3.14 + log(0.542)$
$pH = 3.14 + (-0.266)$
$pH = 2.87$
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(b)
1000ml = 1L
25ml = 0.025 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HNO_2) * V(HNO_2) = 0.132* 0.025 = 3.3 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.116* 0.02 = 2.31 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HNO_2(aq) + NaOH(aq) -- \gt NaNO_2(aq) + H_2O(l)$
- Total volume: 0.025 + 0.02 = 0.045L
3. Since the base is the limiting reactant, only $ 0.00232$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HNO_2] = 0.0033 - 0.00232 = 9.8 \times 10^{-4}$ moles.
Concentration: $\frac{9.8 \times 10^{-4}}{ 0.045} = 0.0218M$
$[NaOH] = 0.00232 - 0.00232 = 0$
$[NaNO_2] = 0 + 0.00232 = 0.00232$ moles.
Concentration: $\frac{ 0.00232}{ 0.045} = 0.0516M$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.14 + log(\frac{0.0516}{0.0218})$
$pH = 3.14 + log(2.37)$
$pH = 3.14 + 0.374$
$pH = 3.51$
