Certainly! Let's solve the given question step-by-step.
1. pH of the Solution:
- The pH of the solution is given as 13.
2. Concentration of [tex]\( \text{OH}^- \)[/tex] Ions:
- To find the concentration of [tex]\( \text{OH}^- \)[/tex] ions, we first need to calculate the pOH using the relationship: [tex]\( \text{pOH} = 14 - \text{pH} \)[/tex].
- For a pH of 13: [tex]\[ \text{pOH} = 14 - 13 = 1 \][/tex]
- The concentration of [tex]\( \text{OH}^- \)[/tex] ions is then found using: [tex]\[ [\text{OH}^-] = 10^{-\text{pOH}} \][/tex]
- Substituting in the pOH value: [tex]\[ [\text{OH}^-] = 10^{-1} = 0.1 \][/tex]
- Thus, the concentration of [tex]\( \text{OH}^- \)[/tex] ions is [tex]\( 0.1 \)[/tex] moles per liter.
3. Concentration of [tex]\( \text{H}^+ \)[/tex] Ions:
- The concentration of [tex]\( \text{H}^+ \)[/tex] ions is related to the pH by: [tex]\[ [\text{H}^+] = 10^{-\text{pH}} \][/tex]
- For a pH of 13: [tex]\[ [\text{H}^+] = 10^{-13} \][/tex]
- Thus, the concentration of [tex]\( \text{H}^+ \)[/tex] ions is [tex]\( 1 \times 10^{-13} \)[/tex] moles per liter.
4. pH of a Different Solution with Known [tex]\( [\text{H}^+] \)[/tex]:
- For a solution with an [tex]\( \text{H}^+ \)[/tex] concentration of [tex]\( 1.0 \times 10^{-4} \)[/tex] moles per liter, we can find the pH using: [tex]\[ \text{pH} = -\log_{10}[\text{H}^+] \][/tex]
- Substituting the given concentration: [tex]\[ \text{pH} = -\log_{10}(1.0 \times 10^{-4}) \][/tex]
- Therefore, the pH is: [tex]\[ \text{pH} = 4.0 \][/tex]
Summary of Solutions:
- A solution with a pH of 13 has approximately [tex]\( 0.1 \)[/tex] moles of [tex]\( \text{OH}^- \)[/tex] ions per liter.
- The same solution has approximately [tex]\( 1 \times 10^{-13} \)[/tex] moles of [tex]\( \text{H}^+ \)[/tex] ions per liter (in decimal form, [tex]\( 0.0000000000001 \)[/tex]).
- A different solution with an [tex]\( \text{H}^+ \)[/tex] concentration of [tex]\( 1.0 \times 10^{-4} \)[/tex] moles per liter has a pH of 4.0.
