Answer :
Let's solve this step-by-step from the given information.
1. Given Information:
- The solution has a [tex]\( \text{pH} \)[/tex] of 4.20.
2. Relationship Between pH and pOH:
- We know that the sum of [tex]\( \text{pH} \)[/tex] and [tex]\( \text{pOH} \)[/tex] is 14.
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]
- Substituting the given [tex]\( \text{pH} \)[/tex]:
[tex]\[ 4.20 + \text{pOH} = 14 \][/tex]
- Solving for [tex]\( \text{pOH} \)[/tex]:
[tex]\[ \text{pOH} = 14 - 4.20 = 9.80 \][/tex]
3. Calculating the Concentration of OH⁻ Ion:
- The concentration of [tex]\( \text{OH}^- \)[/tex] is related to [tex]\( \text{pOH} \)[/tex] by the formula:
[tex]\[ \text{[OH}^-] = 10^{-\text{pOH}} \][/tex]
- Substituting the value of [tex]\( \text{pOH} \)[/tex]:
[tex]\[ \text{[OH}^-] = 10^{-9.80} \][/tex]
- Using exponential calculation (approximated):
[tex]\[ \text{[OH}^-] \approx 1.584893192461111 \times 10^{-10} \, M \][/tex]
4. Select the Closest Answer:
From the given options:
- A. [tex]\( 9.9 \times 10^{-1} \, M \)[/tex]
- B. [tex]\( 6.2 \times 10^{-1} \, M \)[/tex]
- C. [tex]\( 6.3 \times 10^{-5} \, M \)[/tex]
- D. [tex]\( 6.7 \times 10^{-6} \, M \)[/tex]
- E. [tex]\( 1.6 \times 10^{-10} \, M \)[/tex]
The calculated concentration [tex]\(\left(1.584893192461111 \times 10^{-10} \, M\right)\)[/tex] is closest to:
[tex]\[ \text{E. } 1.6 \times 10^{-10} \, M \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{E. \quad 1.6 \times 10^{-10} \, M} \][/tex]
1. Given Information:
- The solution has a [tex]\( \text{pH} \)[/tex] of 4.20.
2. Relationship Between pH and pOH:
- We know that the sum of [tex]\( \text{pH} \)[/tex] and [tex]\( \text{pOH} \)[/tex] is 14.
[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]
- Substituting the given [tex]\( \text{pH} \)[/tex]:
[tex]\[ 4.20 + \text{pOH} = 14 \][/tex]
- Solving for [tex]\( \text{pOH} \)[/tex]:
[tex]\[ \text{pOH} = 14 - 4.20 = 9.80 \][/tex]
3. Calculating the Concentration of OH⁻ Ion:
- The concentration of [tex]\( \text{OH}^- \)[/tex] is related to [tex]\( \text{pOH} \)[/tex] by the formula:
[tex]\[ \text{[OH}^-] = 10^{-\text{pOH}} \][/tex]
- Substituting the value of [tex]\( \text{pOH} \)[/tex]:
[tex]\[ \text{[OH}^-] = 10^{-9.80} \][/tex]
- Using exponential calculation (approximated):
[tex]\[ \text{[OH}^-] \approx 1.584893192461111 \times 10^{-10} \, M \][/tex]
4. Select the Closest Answer:
From the given options:
- A. [tex]\( 9.9 \times 10^{-1} \, M \)[/tex]
- B. [tex]\( 6.2 \times 10^{-1} \, M \)[/tex]
- C. [tex]\( 6.3 \times 10^{-5} \, M \)[/tex]
- D. [tex]\( 6.7 \times 10^{-6} \, M \)[/tex]
- E. [tex]\( 1.6 \times 10^{-10} \, M \)[/tex]
The calculated concentration [tex]\(\left(1.584893192461111 \times 10^{-10} \, M\right)\)[/tex] is closest to:
[tex]\[ \text{E. } 1.6 \times 10^{-10} \, M \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{E. \quad 1.6 \times 10^{-10} \, M} \][/tex]
