The hydroxide ion concentrations will need to be determined to use later in the calculations. The dilution formula is: M₁V₁= = M₂V₂ a. 2.0-mL of 0.10M NaOH and 8.0-mL of water are combined initially. What is the new concentration of the NaOH in the combined solution? (Note the total volume is 10.0- mL) b. 4.0-mL of 0.10M NaOH and 6.0-mL of water are combined initially. What is the new concentration of the NaOH in the combined solution? C. The solution in (a) is combined with 10.0mL of CV. What is the concentration of the NaOH solution? (Note the total volume is now 20.0-mL) d. The solution in (b) is combined with 10.0mL of CV. What is the concentration of the NaOH solution?
The hydroxide ion concentrations will need to be determined to use later in the calculations. The dilution formula is: M₁V₁= = M₂V₂ a. 2.0-mL of 0.10M NaOH and 8.0-mL of water are combined initially. What is the new concentration of the NaOH in the combined solution? (Note the total volume is 10.0- mL) b. 4.0-mL of 0.10M NaOH and 6.0-mL of water are combined initially. What is the new concentration of the NaOH in the combined solution? C. The solution in (a) is combined with 10.0mL of CV. What is the concentration of the NaOH solution? (Note the total volume is now 20.0-mL) d. The solution in (b) is combined with 10.0mL of CV. What is the concentration of the NaOH solution?
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Chapter15: Acids And Bases
Section: Chapter Questions
Problem 15.116QP: Ethanol, CH3CH2OH, can undergo auto-ionization. Write the chemical equation for this...
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**Title: Understanding Hydroxide Ion Concentrations and Rates of Reactions**
**Introduction:**
In this section, we explore how to determine hydroxide ion (OH-) concentrations in solutions through dilution and how to approximate the order of reactions using given rate constants.
**Hydroxide Ion Concentration Calculations:**
The hydroxide ion concentrations will need to be determined to use later in the calculations. The dilution formula is:
\[ M_1V_1 = M_2V_2 \]
**Problems and Solutions:**
**a.** *Concentration in Combined Solutions:*
2.0 mL of 0.10 M NaOH and 8.0 mL of water are combined initially. What is the new concentration of the NaOH in the combined solution? (Note the total volume is 10.0 mL)
**Solution:**
\[ M_1 = 0.10 \text{ M}, V_1 = 2.0 \text{ mL}, V_2 = 10.0 \text{ mL} \]
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{(0.10)(2.0)}{10.0} = 0.02 \text{ M} \]
**b.** *Different Initial Volumes:*
4.0 mL of 0.10 M NaOH and 6.0 mL of water are combined initially. What is the new concentration of the NaOH in the combined solution?
**Solution:**
\[ M_1 = 0.10 \text{ M}, V_1 = 4.0 \text{ mL}, V_2 = 10.0 \text{ mL} \]
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{(0.10)(4.0)}{10.0} = 0.04 \text{ M} \]
**c.** *Further Dilution:*
The solution in (a) is combined with 10.0 mL of CV. What is the concentration of the NaOH solution? (Note the total volume is now 20.0 mL)
**Solution:**
The concentration of NaOH from (a) is 0.02 M with a volume of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85d00984-ae34-4416-a50e-e3da2d4d22d5%2F1aaf1967-6d31-4712-90ae-80a3c9bf7182%2Ftcq758k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
**Title: Understanding Hydroxide Ion Concentrations and Rates of Reactions**
**Introduction:**
In this section, we explore how to determine hydroxide ion (OH-) concentrations in solutions through dilution and how to approximate the order of reactions using given rate constants.
**Hydroxide Ion Concentration Calculations:**
The hydroxide ion concentrations will need to be determined to use later in the calculations. The dilution formula is:
\[ M_1V_1 = M_2V_2 \]
**Problems and Solutions:**
**a.** *Concentration in Combined Solutions:*
2.0 mL of 0.10 M NaOH and 8.0 mL of water are combined initially. What is the new concentration of the NaOH in the combined solution? (Note the total volume is 10.0 mL)
**Solution:**
\[ M_1 = 0.10 \text{ M}, V_1 = 2.0 \text{ mL}, V_2 = 10.0 \text{ mL} \]
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{(0.10)(2.0)}{10.0} = 0.02 \text{ M} \]
**b.** *Different Initial Volumes:*
4.0 mL of 0.10 M NaOH and 6.0 mL of water are combined initially. What is the new concentration of the NaOH in the combined solution?
**Solution:**
\[ M_1 = 0.10 \text{ M}, V_1 = 4.0 \text{ mL}, V_2 = 10.0 \text{ mL} \]
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{(0.10)(4.0)}{10.0} = 0.04 \text{ M} \]
**c.** *Further Dilution:*
The solution in (a) is combined with 10.0 mL of CV. What is the concentration of the NaOH solution? (Note the total volume is now 20.0 mL)
**Solution:**
The concentration of NaOH from (a) is 0.02 M with a volume of
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