(e) Consider this example problem: If 100. mL of 0.100 M HCI solution is mixed with 100. mL of 0.100 M NAOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H20, which is negligible). HCl 1 mol NaOHa -> Nacl H,Om Rxn ratio: 1 mol 1 mol 1 mol (0.100 mmol NaON 1 ml (0.100 mmol NCI Mols @ Start: 100 ml. x 100 ml x O mol www.w. 1ml = 10 mmol HBr = 10 mmol NaOH Change - 10 mmol - 10 mmol NaOH + 10 mmol After rxn O mmol O mmol 10 mmol The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is 10 ттol = 0.05 M NaCl %3D 200 ml Fill in a table like this for the calculations in problem (1) (a)-(d). (2) Now, show the calculations for the samereaction (HBr + NaOH) if 100. mL of 0.250 M HBr and 200.0 mL of 0.250 M NaOH are mixed. Determine the molarities of the solutes in the resulting solution. 4.
(e) Consider this example problem: If 100. mL of 0.100 M HCI solution is mixed with 100. mL of 0.100 M NAOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H20, which is negligible). HCl 1 mol NaOHa -> Nacl H,Om Rxn ratio: 1 mol 1 mol 1 mol (0.100 mmol NaON 1 ml (0.100 mmol NCI Mols @ Start: 100 ml. x 100 ml x O mol www.w. 1ml = 10 mmol HBr = 10 mmol NaOH Change - 10 mmol - 10 mmol NaOH + 10 mmol After rxn O mmol O mmol 10 mmol The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is 10 ттol = 0.05 M NaCl %3D 200 ml Fill in a table like this for the calculations in problem (1) (a)-(d). (2) Now, show the calculations for the samereaction (HBr + NaOH) if 100. mL of 0.250 M HBr and 200.0 mL of 0.250 M NaOH are mixed. Determine the molarities of the solutes in the resulting solution. 4.
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Chapter1: Chemical Foundations
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I am confused on both questions (e) and (2). It is a worksheet about chemistry titration
![Consider this example problem:
If 100. mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (Assume the volumes are additive and ignore the change in H₂O, which is negligible).
\[
\begin{array}{cccccc}
\text{Rxn ratio:} & \text{HCl}_{(aq)} & + & \text{NaOH}_{(aq)} & \rightarrow & \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \\
& 1 \text{ mol} & & 1 \text{ mol} & & 1 \text{ mol} \\
\hline
\text{Mols @ Start:} & 100 \text{ mL} \times \left( \frac{0.100 \text{ mmol HCl}}{1 \text{ mL}} \right) & & 100 \text{ mL} \times \left( \frac{0.100 \text{ mmol NaOH}}{1 \text{ mL}} \right) & & 0 \text{ mol} \\
& = 10 \text{ mmol HBr} & & = 10 \text{ mmol NaOH} & & \\
\text{Change:} & -10 \text{ mmol} & & -10 \text{ mmol} & & +10 \text{ mmol} \\
\hline
\text{After rxn:} & 0 \text{ mmol} & & 0 \text{ mmol} & & 10 \text{ mmol} \\
\end{array}
\]
The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is:
\[
\frac{10 \text{ mmol}}{200 \text{ mL}} = 0.05 \, M \, \text{NaCl}
\]
Fill in a table like this for the calculations in problem (1)(a)-(d).
Now, show the calculations for the same reaction (HBr + NaOH) if 100. mL of 0.250 M HBr and 200.0 mL of 0.250](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdbf8765a-c0ab-46fe-8663-61e338d72b7c%2F37081e8b-49d4-4512-b4d2-e6511765bbf9%2Fikgv1vi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider this example problem:
If 100. mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (Assume the volumes are additive and ignore the change in H₂O, which is negligible).
\[
\begin{array}{cccccc}
\text{Rxn ratio:} & \text{HCl}_{(aq)} & + & \text{NaOH}_{(aq)} & \rightarrow & \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \\
& 1 \text{ mol} & & 1 \text{ mol} & & 1 \text{ mol} \\
\hline
\text{Mols @ Start:} & 100 \text{ mL} \times \left( \frac{0.100 \text{ mmol HCl}}{1 \text{ mL}} \right) & & 100 \text{ mL} \times \left( \frac{0.100 \text{ mmol NaOH}}{1 \text{ mL}} \right) & & 0 \text{ mol} \\
& = 10 \text{ mmol HBr} & & = 10 \text{ mmol NaOH} & & \\
\text{Change:} & -10 \text{ mmol} & & -10 \text{ mmol} & & +10 \text{ mmol} \\
\hline
\text{After rxn:} & 0 \text{ mmol} & & 0 \text{ mmol} & & 10 \text{ mmol} \\
\end{array}
\]
The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is:
\[
\frac{10 \text{ mmol}}{200 \text{ mL}} = 0.05 \, M \, \text{NaCl}
\]
Fill in a table like this for the calculations in problem (1)(a)-(d).
Now, show the calculations for the same reaction (HBr + NaOH) if 100. mL of 0.250 M HBr and 200.0 mL of 0.250
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