Answer :
Let's go through the steps to calculate the mass of sodium bicarbonate in the antacid tablets.
### 1. Determine the Volume of HCl Added
We begin by finding the volume of hydrochloric acid (HCl) added for each trial. This is calculated by subtracting the initial burette reading from the final burette reading for each trial.
- Trial 1: [tex]\(11.6 \, mL - 0.6 \, mL = 11.0 \, mL\)[/tex]
- Trial 2: [tex]\(21.7 \, mL - 11.6 \, mL = 10.1 \, mL\)[/tex]
- Trial 3: [tex]\(31.7 \, mL - 21.7 \, mL = 10.0 \, mL\)[/tex]
- Trial 4: [tex]\(41.8 \, mL - 31.7 \, mL = 10.1 \, mL\)[/tex]
So, the volumes of HCl added are:
[tex]\[ [11.0 \, mL, 10.1 \, mL, 10.0 \, mL, 10.1 \, mL] \][/tex]
### 2. Calculate the Moles of HCl Added
Next, we use the concentration of HCl to find the number of moles of HCl added in each trial. The concentration of HCl is given as [tex]\(0.200 \, mol/L\)[/tex].
To convert from mL to L and then find the moles, we use the formula:
[tex]\[ \text{Moles of HCl} = \text{Volume of HCl} \times \text{Concentration of HCl} \][/tex]
- Trial 1: [tex]\(11.0 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.0022 \, mol\)[/tex]
- Trial 2: [tex]\(10.1 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.00202 \, mol\)[/tex]
- Trial 3: [tex]\(10.0 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.0020 \, mol\)[/tex]
- Trial 4: [tex]\(10.1 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.00202 \, mol\)[/tex]
So, the moles of HCl added are:
[tex]\[ [0.0022 \, mol, 0.00202 \, mol, 0.0020 \, mol, 0.00202 \, mol] \][/tex]
### 3. Determine the Moles of NaHCO₃
Given that sodium bicarbonate (NaHCO₃) reacts in a 1:1 molar ratio with HCl, the moles of NaHCO₃ will be equal to the moles of HCl added.
So, the moles of NaHCO₃ are:
[tex]\[ [0.0022 \, mol, 0.00202 \, mol, 0.0020 \, mol, 0.00202 \, mol] \][/tex]
### 4. Calculate the Mass of NaHCO₃
To find the mass of sodium bicarbonate in each trial, we use the molar mass of NaHCO₃, which is [tex]\(84.007 \, g/mol\)[/tex].
[tex]\[ \text{Mass of NaHCO₃} = \text{Moles of NaHCO₃} \times \text{Molar mass of NaHCO₃} \][/tex]
Converting the mass to mg by multiplying by [tex]\(1000 \, mg/g\)[/tex],
- Trial 1: [tex]\(0.0022 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 184.8154 \, mg\)[/tex]
- Trial 2: [tex]\(0.00202 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 169.69414 \, mg\)[/tex]
- Trial 3: [tex]\(0.0020 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 168.014 \, mg\)[/tex]
- Trial 4: [tex]\(0.00202 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 169.69414 \, mg\)[/tex]
So, the masses of NaHCO₃ in each tablet are:
[tex]\[ [184.8154 \, mg, 169.69414 \, mg, 168.014 \, mg, 169.69414 \, mg] \][/tex]
### Conclusion
The calculated masses of sodium bicarbonate in the antacid tablets for the trials are:
[tex]\[ \boxed{184.8154 \, mg, 169.69414 \, mg, 168.014 \, mg, 169.69414 \, mg} \][/tex]
Based on the calculated values, the manufacturer's claim that each antacid tablet contains 200 mg of sodium bicarbonate does not hold true for these trials, as the average mass is significantly lower.
### 1. Determine the Volume of HCl Added
We begin by finding the volume of hydrochloric acid (HCl) added for each trial. This is calculated by subtracting the initial burette reading from the final burette reading for each trial.
- Trial 1: [tex]\(11.6 \, mL - 0.6 \, mL = 11.0 \, mL\)[/tex]
- Trial 2: [tex]\(21.7 \, mL - 11.6 \, mL = 10.1 \, mL\)[/tex]
- Trial 3: [tex]\(31.7 \, mL - 21.7 \, mL = 10.0 \, mL\)[/tex]
- Trial 4: [tex]\(41.8 \, mL - 31.7 \, mL = 10.1 \, mL\)[/tex]
So, the volumes of HCl added are:
[tex]\[ [11.0 \, mL, 10.1 \, mL, 10.0 \, mL, 10.1 \, mL] \][/tex]
### 2. Calculate the Moles of HCl Added
Next, we use the concentration of HCl to find the number of moles of HCl added in each trial. The concentration of HCl is given as [tex]\(0.200 \, mol/L\)[/tex].
To convert from mL to L and then find the moles, we use the formula:
[tex]\[ \text{Moles of HCl} = \text{Volume of HCl} \times \text{Concentration of HCl} \][/tex]
- Trial 1: [tex]\(11.0 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.0022 \, mol\)[/tex]
- Trial 2: [tex]\(10.1 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.00202 \, mol\)[/tex]
- Trial 3: [tex]\(10.0 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.0020 \, mol\)[/tex]
- Trial 4: [tex]\(10.1 \, mL \times 0.200 \, mol/L \times \frac{1 \, L}{1000 \, mL} = 0.00202 \, mol\)[/tex]
So, the moles of HCl added are:
[tex]\[ [0.0022 \, mol, 0.00202 \, mol, 0.0020 \, mol, 0.00202 \, mol] \][/tex]
### 3. Determine the Moles of NaHCO₃
Given that sodium bicarbonate (NaHCO₃) reacts in a 1:1 molar ratio with HCl, the moles of NaHCO₃ will be equal to the moles of HCl added.
So, the moles of NaHCO₃ are:
[tex]\[ [0.0022 \, mol, 0.00202 \, mol, 0.0020 \, mol, 0.00202 \, mol] \][/tex]
### 4. Calculate the Mass of NaHCO₃
To find the mass of sodium bicarbonate in each trial, we use the molar mass of NaHCO₃, which is [tex]\(84.007 \, g/mol\)[/tex].
[tex]\[ \text{Mass of NaHCO₃} = \text{Moles of NaHCO₃} \times \text{Molar mass of NaHCO₃} \][/tex]
Converting the mass to mg by multiplying by [tex]\(1000 \, mg/g\)[/tex],
- Trial 1: [tex]\(0.0022 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 184.8154 \, mg\)[/tex]
- Trial 2: [tex]\(0.00202 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 169.69414 \, mg\)[/tex]
- Trial 3: [tex]\(0.0020 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 168.014 \, mg\)[/tex]
- Trial 4: [tex]\(0.00202 \, mol \times 84.007 \, g/mol \times 1000 \, mg/g = 169.69414 \, mg\)[/tex]
So, the masses of NaHCO₃ in each tablet are:
[tex]\[ [184.8154 \, mg, 169.69414 \, mg, 168.014 \, mg, 169.69414 \, mg] \][/tex]
### Conclusion
The calculated masses of sodium bicarbonate in the antacid tablets for the trials are:
[tex]\[ \boxed{184.8154 \, mg, 169.69414 \, mg, 168.014 \, mg, 169.69414 \, mg} \][/tex]
Based on the calculated values, the manufacturer's claim that each antacid tablet contains 200 mg of sodium bicarbonate does not hold true for these trials, as the average mass is significantly lower.
