Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.