Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

17. Acid and Base Equilibrium

pH of Strong Acids and Bases

General Chemistry

17. Acid and Base Equilibrium

pH of Strong Acids and Bases

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Multiple Choice

What is the pH of a solution made by mixing 25.00 mL of 0.100 M HCl with 40.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive.

pH = 12.00

pH = 5.00

pH = 2.00

pH = 7.00

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Verified step by step guidance

First, determine the moles of HCl and KOH in the solution. Use the formula \( \text{moles} = \text{concentration} \times \text{volume} \). For HCl: \( \text{moles of HCl} = 0.100 \, \text{M} \times 25.00 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \). For KOH: \( \text{moles of KOH} = 0.100 \, \text{M} \times 40.00 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \).

Next, identify the reaction between HCl and KOH. The balanced chemical equation is \( \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \). This is a neutralization reaction where HCl and KOH react in a 1:1 molar ratio.

Calculate the moles of HCl and KOH that react. Since they react in a 1:1 ratio, the limiting reactant will be the one with fewer moles. Compare the moles of HCl and KOH calculated in the first step to determine the limiting reactant.

Determine the moles of excess KOH after the reaction. Subtract the moles of HCl from the moles of KOH to find the moles of KOH remaining after the reaction.

Calculate the concentration of the excess KOH in the final solution. Use the formula \( \text{concentration} = \frac{\text{moles of excess KOH}}{\text{total volume of solution in liters}} \). The total volume is the sum of the initial volumes of HCl and KOH solutions. Finally, use the concentration of KOH to find the pOH and then the pH using the relation \( \text{pH} + \text{pOH} = 14 \).