A solid mixture weighing $5 \text{ g}$ contains equal number of moles of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$. This solid mixture was dissolved in $1 \text{ L}$ of water. What is the volume (in $\text{mL}$) of $0.1 \text{ M } \text{HCl}$ required to completely react with this $1 \text{ L}$ mixture solution?
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For acid-base titrations, the key step is to balance the reaction equations to find the correct stoichiometric ratio (moles of acid per mole of base). The volume of titrant is found using the formula $V_{\text{acid}} = n_{\text{acid}} / M_{\text{acid}}$.
Step 1: Calculate the molar mass of the components.
Atomic masses: $\text{Na}=23, \text{C}=12, \text{O}=16, \text{H}=1$.
Molar mass of $\text{Na}_2\text{CO}_3$: $2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 \text{ g/mol}$.
Molar mass of $\text{NaHCO}_3$: $23 + 1 + 12 + 3(16) = 84 \text{ g/mol}$.
Step 2: Calculate the number of moles of each component.
Let $n$ be the number of moles of each component (since they are present in equal moles).
The total mass of the mixture is $W = 5 \text{ g}$.
\[
W = n \cdot M_{\text{Na}_2\text{CO}_3} + n \cdot M_{\text{NaHCO}_3} = n(106) + n(84).
\]
\[
5 \text{ g} = n(106 + 84) = 190n.
\]
\[
n = \frac{5}{190} = \frac{1}{38} \text{ moles}.
\]
Step 3: Write the reaction of each component with $\text{HCl$.}
Both sodium carbonate and sodium bicarbonate are bases and react with the acid $\text{HCl}$.
Reaction 1: $\text{Na}_2\text{CO}_3$ (dibasic) reacts with 2 moles of $\text{HCl}$.
\[
\text{Na}_2\text{CO}_3 + 2\text{HCl} \to 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2.
\]
Reaction 2: $\text{NaHCO}_3$ (monobasic) reacts with 1 mole of $\text{HCl}$.
\[
\text{NaHCO}_3 + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2.
\]
Step 4: Calculate the total moles of $\text{HCl$ required.}
Moles of $\text{HCl}$ required for $\text{Na}_2\text{CO}_3$: $2 \times n = 2 \times \frac{1}{38} = \frac{2}{38} \text{ moles}$.
Moles of $\text{HCl}$ required for $\text{NaHCO}_3$: $1 \times n = 1 \times \frac{1}{38} = \frac{1}{38} \text{ moles}$.
Total moles of $\text{HCl}$ required: $n_{\text{HCl}} = \frac{2}{38} + \frac{1}{38} = \frac{3}{38} \text{ moles}$.
Step 5: Calculate the volume of $\text{HCl$ solution required.}
We use the formula $V_{\text{HCl}} = \frac{n_{\text{HCl}}}{M_{\text{HCl}}}$.
We are given $M_{\text{HCl}} = 0.1 \text{ M}$.
\[
V_{\text{HCl}} (\text{L}) = \frac{3/38 \text{ moles}}{0.1 \text{ mol/L}} = \frac{3}{38 \times 0.1} \text{ L} = \frac{3}{3.8} \text{ L}.
\]
$V_{\text{HCl}} (\text{L}) \approx 0.78947 \text{ L}$.
Convert to $\text{mL}$:
\[
V_{\text{HCl}} (\text{mL}) = 0.78947 \times 1000 \text{ mL} \approx 789.47 \text{ mL}.
\]
The volume is approximately $789.0 \text{ mL}$.
