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Sagot :
To calculate the molarity of the [tex]\(\text{H}_2\text{SO}_4\)[/tex] solution for each titration, we can use the equation derived from the neutralization reaction:
[tex]\[ M_A \times V_A = M_B \times V_B \][/tex]
where:
- [tex]\( M_A \)[/tex] is the molarity of the acid [tex]\(\text{H}_2\text{SO}_4\)[/tex] (which we need to find),
- [tex]\( V_A \)[/tex] is the volume of the acid in mL,
- [tex]\( M_B \)[/tex] is the molarity of the base (given in the table),
- [tex]\( V_B \)[/tex] is the volume of the base in mL (given in the table).
To calculate [tex]\( M_A \)[/tex], we rearrange the formula:
[tex]\[ M_A = \frac{M_B \times V_B}{V_A} \][/tex]
Given values from the table:
- [tex]\( V_A = 20 \, \text{mL} \)[/tex] for all titrations.
For Titration 1:
- [tex]\( V_B = 0.15 \, \text{mL} \)[/tex]
- [tex]\( M_B = 18.20 \)[/tex]
Thus,
[tex]\[ M_A = \frac{18.20 \times 0.15}{20} = \frac{2.73}{20} = 0.1365 \][/tex]
For Titration 2:
- [tex]\( V_B = 0.70 \, \text{mL} \)[/tex]
- [tex]\( M_B = 18.60 \)[/tex]
Thus,
[tex]\[ M_A = \frac{18.60 \times 0.70}{20} = \frac{13.02}{20} = 0.651 \][/tex]
For Titration 3:
- [tex]\( V_B = 18.30 \, \text{mL} \)[/tex]
- [tex]\( M_B = 34.40 \)[/tex]
Thus,
[tex]\[ M_A = \frac{34.40 \times 18.30}{20} = \frac{629.52}{20} = 31.476 \][/tex]
So the molarity of the [tex]\(\text{H}_2\text{SO}_4\)[/tex] solutions for each titration are:
1. Titration 1: [tex]\( M_A = 0.1365 \)[/tex]
2. Titration 2: [tex]\( M_A = 0.651 \)[/tex]
3. Titration 3: [tex]\( M_A = 31.476 \)[/tex]
The completed table is:
[tex]\[ \begin{array}{|l|l|l|l|} \hline & \text{\textbf{Titration 1}} & \text{\textbf{Titration 2}} & \text{\textbf{Titration 3}} \\ \hline \text{Volume of Acid} \, (V_A) \, \text{(in mL)} & 20 \, \text{mL} & 20 \, \text{mL} & 20 \, \text{mL} \\ \hline \text{Volume of Base} \, (V_B) \, \text{(in mL)} & 0.15 & 0.70 & 18.30 \\ \hline \text{Molarity of Base} \, (M_B) & 18.20 & 18.60 & 34.40 \\ \hline \text{Molarity of Acid} \, (M_A) & 0.1365 & 0.651 & 31.476 \\ \hline \end{array} \][/tex]
These molarities represent the concentration of [tex]\(\text{H}_2\text{SO}_4\)[/tex] for each of the three titrations.
[tex]\[ M_A \times V_A = M_B \times V_B \][/tex]
where:
- [tex]\( M_A \)[/tex] is the molarity of the acid [tex]\(\text{H}_2\text{SO}_4\)[/tex] (which we need to find),
- [tex]\( V_A \)[/tex] is the volume of the acid in mL,
- [tex]\( M_B \)[/tex] is the molarity of the base (given in the table),
- [tex]\( V_B \)[/tex] is the volume of the base in mL (given in the table).
To calculate [tex]\( M_A \)[/tex], we rearrange the formula:
[tex]\[ M_A = \frac{M_B \times V_B}{V_A} \][/tex]
Given values from the table:
- [tex]\( V_A = 20 \, \text{mL} \)[/tex] for all titrations.
For Titration 1:
- [tex]\( V_B = 0.15 \, \text{mL} \)[/tex]
- [tex]\( M_B = 18.20 \)[/tex]
Thus,
[tex]\[ M_A = \frac{18.20 \times 0.15}{20} = \frac{2.73}{20} = 0.1365 \][/tex]
For Titration 2:
- [tex]\( V_B = 0.70 \, \text{mL} \)[/tex]
- [tex]\( M_B = 18.60 \)[/tex]
Thus,
[tex]\[ M_A = \frac{18.60 \times 0.70}{20} = \frac{13.02}{20} = 0.651 \][/tex]
For Titration 3:
- [tex]\( V_B = 18.30 \, \text{mL} \)[/tex]
- [tex]\( M_B = 34.40 \)[/tex]
Thus,
[tex]\[ M_A = \frac{34.40 \times 18.30}{20} = \frac{629.52}{20} = 31.476 \][/tex]
So the molarity of the [tex]\(\text{H}_2\text{SO}_4\)[/tex] solutions for each titration are:
1. Titration 1: [tex]\( M_A = 0.1365 \)[/tex]
2. Titration 2: [tex]\( M_A = 0.651 \)[/tex]
3. Titration 3: [tex]\( M_A = 31.476 \)[/tex]
The completed table is:
[tex]\[ \begin{array}{|l|l|l|l|} \hline & \text{\textbf{Titration 1}} & \text{\textbf{Titration 2}} & \text{\textbf{Titration 3}} \\ \hline \text{Volume of Acid} \, (V_A) \, \text{(in mL)} & 20 \, \text{mL} & 20 \, \text{mL} & 20 \, \text{mL} \\ \hline \text{Volume of Base} \, (V_B) \, \text{(in mL)} & 0.15 & 0.70 & 18.30 \\ \hline \text{Molarity of Base} \, (M_B) & 18.20 & 18.60 & 34.40 \\ \hline \text{Molarity of Acid} \, (M_A) & 0.1365 & 0.651 & 31.476 \\ \hline \end{array} \][/tex]
These molarities represent the concentration of [tex]\(\text{H}_2\text{SO}_4\)[/tex] for each of the three titrations.
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