Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Fill out the following table and answer questions 1 and 2.

\begin{tabular}{|l|l|l|l|}
\hline & Titration 1 & Titration 2 & Titration 3 \\
\hline \begin{tabular}{c}
Volume of Acid [tex]$\left( V _{ A }\right)$[/tex] \\
(in mL )
\end{tabular} & 20 mL & 20 mL & 20 mL \\
\hline \begin{tabular}{c}
Volume of Base [tex]$\left( V _{ B }\right)$[/tex] \\
(in mL )
\end{tabular} & 0.15 & 0.70 & 18.30 \\
\hline \begin{tabular}{l}
Molarity of Base [tex]$\left( M _{ B }\right)$[/tex]
\end{tabular} & 18.20 & 18.60 & 34.40 \\
\hline
\end{tabular}

1. Calculate the molarity of the [tex]$H_2SO_4$[/tex] solution for each of the three runs.

Sagot :

GinnyAnswer
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.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.