Question

A 0.754 g sample containing sodium bicarbonate was dissolved and titrated with standard 0.15 M HCl solution, requiring 38.5 ml. Calculate the percent sodium bicarbonate in the sample. Also write chemical reaction.

Answer 1

[AI-generated answer — Bot GPT-4o]

To calculate the percent sodium bicarbonate in the sample and write the chemical reaction, we need to follow these steps:

### Step 1: Write the chemical reaction
The reaction between sodium bicarbonate (NaHCO₃) and hydrochloric acid (HCl) is as follows:

\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate the moles of HCl used in the titration
We are given:
- Concentration of HCl solution (\(M_HCl\)) = 0.15 M
- Volume of HCl solution used (\(V_HCl\)) = 38.5 ml = 0.0385 L

\[ \text{Moles of HCl} = \text{M}_\text{HCl} \times \text{V}_\text{HCl} \]
\[ \text{Moles of HCl} = 0.15 \, M \times 0.0385 \, L \]
\[ \text{Moles of HCl} = 0.005775 \, \text{mol} \]

### Step 3: Use stoichiometry to find moles of NaHCO₃
From the balanced equation, we see that the reaction ratio of NaHCO₃ to HCl is 1:1. Therefore, the moles of NaHCO₃ are the same as the moles of HCl.

\[ \text{Moles of NaHCO}_3 = \text{Moles of HCl} = 0.005775 \, \text{mol} \]

### Step 4: Calculate the mass of NaHCO₃
The molar mass of NaHCO₃ is:

\[
\text{Molar mass of NaHCO}_3 = 23 \, (\text{Na}) + 1 \, (\text{H}) + 12 \, (\text{C}) + 3 \times 16 \, (\text{O})
\]
\[ \text{Molar mass of NaHCO}_3 = 23 + 1 + 12 + 48 = 84 \, \text{g/mol} \]

Now calculate the mass of NaHCO₃:

\[
\text{Mass of NaHCO}_3 = \text{Moles of NaHCO}_3 \times \text{Molar mass of NaHCO}_3
\]
\[
\text{Mass of NaHCO}_3 = 0.005775 \, \text{mol} \times 84 \, \text{g/mol}
\]
\[
\text{Mass of NaHCO}_3 = 0.4851 \, \text{g}
\]

### Step 5: Calculate the percent sodium bicarbonate in the sample
We are given that the total mass of the sample is 0.754 g.

\[
\text{Percent NaHCO}_3 = \left( \frac{\text{Mass of NaHCO}_3}{\text{Total mass of sample}} \right) \times 100
\]
\[
\text{Percent NaHCO}_3 = \left( \frac{0.4851 \, \text{g}}{0.754 \, \text{g}} \right) \times 100
\]
\[
\text{Percent NaHCO}_3 \approx 64.36 \%
\]

### Final answer:
The percent of sodium bicarbonate in the sample is approximately \(64.36\%\). The chemical reaction is:

\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]