Given data: The concentration of hydrofluoric acid = 0.25 M The volume of hydrofluoric acid = 250 mL The pH of the solution is 3.50 Mole calculation: We know that, pk = -log K₁ The K₂ for hydrofluoric acid is 6.8 x 10-4 The moles of hydrofluorie acid that is needed can be calculated using the Henderson-Hasselbalch equation as follows, pH= pk + log- [conjugate base] [weak acid] [Buffer. base] pH= pk + log- [Buffer. acid] [F] 3.50= -log K + log (HF) = -log(6.8 x 10-¹) + log- = 3.167 +log[0.25] -log[HF] log[HF]= 3.167 -0.602 - 3.50 -0.9351 [HF] = 0.1161 M Therefore, the concentration of hydrofluoric acid is 0.1161 M The mole of hydrofluoric acid is calculated as follows, Concentration=; mole Total volume (0.25 M) [HF] How did mol HF=Concentration x Total volume = 0.1161 x 0.250 L = 0.029 mol Therefore, the number of moles required is 0.029 mol they get this?
Given data: The concentration of hydrofluoric acid = 0.25 M The volume of hydrofluoric acid = 250 mL The pH of the solution is 3.50 Mole calculation: We know that, pk = -log K₁ The K₂ for hydrofluoric acid is 6.8 x 10-4 The moles of hydrofluorie acid that is needed can be calculated using the Henderson-Hasselbalch equation as follows, pH= pk + log- [conjugate base] [weak acid] [Buffer. base] pH= pk + log- [Buffer. acid] [F] 3.50= -log K + log (HF) = -log(6.8 x 10-¹) + log- = 3.167 +log[0.25] -log[HF] log[HF]= 3.167 -0.602 - 3.50 -0.9351 [HF] = 0.1161 M Therefore, the concentration of hydrofluoric acid is 0.1161 M The mole of hydrofluoric acid is calculated as follows, Concentration=; mole Total volume (0.25 M) [HF] How did mol HF=Concentration x Total volume = 0.1161 x 0.250 L = 0.029 mol Therefore, the number of moles required is 0.029 mol they get this?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Given Data:**
- The concentration of hydrofluoric acid = 0.25 M
- The volume of hydrofluoric acid = 250 mL
- The pH of the solution is 3.50
**Mole Calculation:**
We know that,
\[ \text{pK}_a = -\log K_a \]
The \( K_a \) for hydrofluoric acid is \( 6.8 \times 10^{-4} \).
The moles of hydrofluoric acid needed can be calculated using the Henderson-Hasselbalch equation as follows:
\[ \text{pH} = \text{pK}_a + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \]
\[ \text{pH} = \text{pK}_a + \log \left(\frac{[\text{Buffer base}]}{[\text{Buffer acid}]}\right) \]
\[ 3.50 = -\log(6.8 \times 10^{-4}) + \log \left(\frac{0.25 \, \text{M}}{[\text{HF}]}\right) \]
\[ 3.167 + \log[0.25] - \log[\text{HF}] \]
\[ \log[\text{HF}] = 3.167 - 0.602 - 3.50 \]
\[ = -0.9351 \]
\[ [\text{HF}] = 0.1161 \, \text{M} \]
Therefore, the concentration of hydrofluoric acid is 0.1161 M.
The mole of hydrofluoric acid is calculated as follows:
\[
\text{Concentration} = \frac{\text{mole}}{\text{Total volume}}
\]
\[
\text{mol HF} = \text{Concentration} \times \text{Total volume}
\]
\[
= 0.1161 \times 0.250 \, \text{L}
\]
\[
= 0.029 \, \text{mol}
\]
Therefore, the number of moles required is **0.029 mol**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6cc73a6-72c8-4e2b-a22b-5b06a5805dfc%2Fbd73ab52-a7dd-4a46-bb15-badeed640d9b%2Fapppt8b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Given Data:**
- The concentration of hydrofluoric acid = 0.25 M
- The volume of hydrofluoric acid = 250 mL
- The pH of the solution is 3.50
**Mole Calculation:**
We know that,
\[ \text{pK}_a = -\log K_a \]
The \( K_a \) for hydrofluoric acid is \( 6.8 \times 10^{-4} \).
The moles of hydrofluoric acid needed can be calculated using the Henderson-Hasselbalch equation as follows:
\[ \text{pH} = \text{pK}_a + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \]
\[ \text{pH} = \text{pK}_a + \log \left(\frac{[\text{Buffer base}]}{[\text{Buffer acid}]}\right) \]
\[ 3.50 = -\log(6.8 \times 10^{-4}) + \log \left(\frac{0.25 \, \text{M}}{[\text{HF}]}\right) \]
\[ 3.167 + \log[0.25] - \log[\text{HF}] \]
\[ \log[\text{HF}] = 3.167 - 0.602 - 3.50 \]
\[ = -0.9351 \]
\[ [\text{HF}] = 0.1161 \, \text{M} \]
Therefore, the concentration of hydrofluoric acid is 0.1161 M.
The mole of hydrofluoric acid is calculated as follows:
\[
\text{Concentration} = \frac{\text{mole}}{\text{Total volume}}
\]
\[
\text{mol HF} = \text{Concentration} \times \text{Total volume}
\]
\[
= 0.1161 \times 0.250 \, \text{L}
\]
\[
= 0.029 \, \text{mol}
\]
Therefore, the number of moles required is **0.029 mol**.
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