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Sagot :
Alright, let's classify each aqueous solution as acidic, basic, or neutral. We will use the pH scale, where pH values less than 7 indicate acidity, a pH of 7 indicates neutrality, and pH values greater than 7 indicate basicity.
1. [tex]$\left[H^{+}\right] = 8.0 \times 10^{-2}$[/tex]:
- The concentration of hydrogen ions, [tex]\( [H^+] \)[/tex], is [tex]\( 8.0 \times 10^{-2} \)[/tex].
- To find the pH, we use the formula [tex]\( \text{pH} = -\log_{10}[H^+] \)[/tex].
- Substituting the given [tex]\( [H^+] \)[/tex], the pH is approximately [tex]\( 1.10 \)[/tex] (note that [tex]\( -\log_{10}(8.0 \times 10^{-2}) = 1.10 \)[/tex]).
- Since pH [tex]\( 1.10 < 7 \)[/tex], the solution is Acidic.
2. pH = 3.05:
- The pH is given directly as 3.05.
- Since pH [tex]\( 3.05 < 7 \)[/tex], the solution is Acidic.
3. [tex]$\left[H^{+}\right] = 1.0 \times 10^{-7}$[/tex]:
- The concentration of hydrogen ions, [tex]\( [H^+] \)[/tex], is [tex]\( 1.0 \times 10^{-7} \)[/tex].
- To find the pH, we use the formula [tex]\( \text{pH} = -\log_{10}[H^+] \)[/tex].
- Substituting the given [tex]\( [H^+] \)[/tex], the pH is approximately [tex]\( 7.00 \)[/tex] (note that [tex]\( -\log_{10}(1.0 \times 10^{-7}) = 7.00 \)[/tex]).
- Since pH [tex]\( 7.00 = 7 \)[/tex], the solution is Neutral.
4. pH = 7.00:
- The pH is given directly as 7.00.
- Since pH [tex]\( 7.00 = 7 \)[/tex], the solution is Neutral.
5. [tex]$\left[H^{+} \right] = 5.3 \times 10^{-8}$[/tex]:
- The concentration of hydrogen ions, [tex]\( [H^+] \)[/tex], is [tex]\( 5.3 \times 10^{-8} \)[/tex].
- To find the pH, we use the formula [tex]\( \text{pH} = -\log_{10}[H^+] \)[/tex].
- Substituting the given [tex]\( [H^+] \)[/tex], the pH is approximately [tex]\( 7.28 \)[/tex] (note that [tex]\( -\log_{10}(5.3 \times 10^{-8}) = 7.28 \)[/tex]).
- Since pH [tex]\( 7.28 > 7 \)[/tex], the solution is Basic.
6. [tex]$\left[OH^{-}\right] = 3.7 \times 10^{-5}$[/tex]:
- The concentration of hydroxide ions, [tex]\( [OH^-] \)[/tex], is [tex]\( 3.7 \times 10^{-5} \)[/tex].
- To find the pH, we first find the pOH using [tex]\( \text{pOH} = -\log_{10}[OH^-] \)[/tex].
- Substituting the given [tex]\( [OH^-] \)[/tex], the pOH is approximately [tex]\( 4.43 \)[/tex] (note that [tex]\( -\log_{10}(3.7 \times 10^{-5}) = 4.43 \)[/tex]).
- Then we use [tex]\( \text{pH} = 14 - \text{pOH} \)[/tex].
- So, [tex]\( \text{pH} = 14 - 4.43 = 9.57 \)[/tex].
- Since pH [tex]\( 9.57 > 7 \)[/tex], the solution is Basic.
7. pH = 8.92:
- The pH is given directly as 8.92.
- Since pH [tex]\( 8.92 > 7 \)[/tex], the solution is Basic.
8. [tex]$\left[OH^{-}\right] = 6.6 \times 10^{-13}$[/tex]:
- The concentration of hydroxide ions, [tex]\( [OH^-] \)[/tex], is [tex]\( 6.6 \times 10^{-13} \)[/tex].
- To find the pH, we first find the pOH using [tex]\( \text{pOH} = -\log_{10}[OH^-] \)[/tex].
- Substituting the given [tex]\( [OH^-] \)[/tex], the pOH is approximately [tex]\( 12.18 \)[/tex] (note that [tex]\( -\log_{10}(6.6 \times 10^{-13}) = 12.18 \)[/tex]).
- Then we use [tex]\( \text{pH} = 14 - \text{pOH} \)[/tex].
- So, [tex]\( \text{pH} = 14 - 12.18 = 1.82 \)[/tex].
- Since pH [tex]\( 1.82 < 7 \)[/tex], the solution is Acidic.
To summarize:
- [tex]$\left[H^{+}\right] = 8.0 \times 10^{-2}$[/tex]: Acidic
- pH = 3.05: Acidic
- [tex]$\left[H^{+}\right] = 1.0 \times 10^{-7}$[/tex]: Neutral
- pH = 7.00: Neutral
- [tex]$\left[H^{+}\right] = 5.3 \times 10^{-8}$[/tex]: Basic
- [tex]$\left[OH^{-}\right] = 3.7 \times 10^{-5}$[/tex]: Basic
- pH = 8.92: Basic
- [tex]$\left[OH^{-}\right] = 6.6 \times 10^{-13}$[/tex]: Acidic
1. [tex]$\left[H^{+}\right] = 8.0 \times 10^{-2}$[/tex]:
- The concentration of hydrogen ions, [tex]\( [H^+] \)[/tex], is [tex]\( 8.0 \times 10^{-2} \)[/tex].
- To find the pH, we use the formula [tex]\( \text{pH} = -\log_{10}[H^+] \)[/tex].
- Substituting the given [tex]\( [H^+] \)[/tex], the pH is approximately [tex]\( 1.10 \)[/tex] (note that [tex]\( -\log_{10}(8.0 \times 10^{-2}) = 1.10 \)[/tex]).
- Since pH [tex]\( 1.10 < 7 \)[/tex], the solution is Acidic.
2. pH = 3.05:
- The pH is given directly as 3.05.
- Since pH [tex]\( 3.05 < 7 \)[/tex], the solution is Acidic.
3. [tex]$\left[H^{+}\right] = 1.0 \times 10^{-7}$[/tex]:
- The concentration of hydrogen ions, [tex]\( [H^+] \)[/tex], is [tex]\( 1.0 \times 10^{-7} \)[/tex].
- To find the pH, we use the formula [tex]\( \text{pH} = -\log_{10}[H^+] \)[/tex].
- Substituting the given [tex]\( [H^+] \)[/tex], the pH is approximately [tex]\( 7.00 \)[/tex] (note that [tex]\( -\log_{10}(1.0 \times 10^{-7}) = 7.00 \)[/tex]).
- Since pH [tex]\( 7.00 = 7 \)[/tex], the solution is Neutral.
4. pH = 7.00:
- The pH is given directly as 7.00.
- Since pH [tex]\( 7.00 = 7 \)[/tex], the solution is Neutral.
5. [tex]$\left[H^{+} \right] = 5.3 \times 10^{-8}$[/tex]:
- The concentration of hydrogen ions, [tex]\( [H^+] \)[/tex], is [tex]\( 5.3 \times 10^{-8} \)[/tex].
- To find the pH, we use the formula [tex]\( \text{pH} = -\log_{10}[H^+] \)[/tex].
- Substituting the given [tex]\( [H^+] \)[/tex], the pH is approximately [tex]\( 7.28 \)[/tex] (note that [tex]\( -\log_{10}(5.3 \times 10^{-8}) = 7.28 \)[/tex]).
- Since pH [tex]\( 7.28 > 7 \)[/tex], the solution is Basic.
6. [tex]$\left[OH^{-}\right] = 3.7 \times 10^{-5}$[/tex]:
- The concentration of hydroxide ions, [tex]\( [OH^-] \)[/tex], is [tex]\( 3.7 \times 10^{-5} \)[/tex].
- To find the pH, we first find the pOH using [tex]\( \text{pOH} = -\log_{10}[OH^-] \)[/tex].
- Substituting the given [tex]\( [OH^-] \)[/tex], the pOH is approximately [tex]\( 4.43 \)[/tex] (note that [tex]\( -\log_{10}(3.7 \times 10^{-5}) = 4.43 \)[/tex]).
- Then we use [tex]\( \text{pH} = 14 - \text{pOH} \)[/tex].
- So, [tex]\( \text{pH} = 14 - 4.43 = 9.57 \)[/tex].
- Since pH [tex]\( 9.57 > 7 \)[/tex], the solution is Basic.
7. pH = 8.92:
- The pH is given directly as 8.92.
- Since pH [tex]\( 8.92 > 7 \)[/tex], the solution is Basic.
8. [tex]$\left[OH^{-}\right] = 6.6 \times 10^{-13}$[/tex]:
- The concentration of hydroxide ions, [tex]\( [OH^-] \)[/tex], is [tex]\( 6.6 \times 10^{-13} \)[/tex].
- To find the pH, we first find the pOH using [tex]\( \text{pOH} = -\log_{10}[OH^-] \)[/tex].
- Substituting the given [tex]\( [OH^-] \)[/tex], the pOH is approximately [tex]\( 12.18 \)[/tex] (note that [tex]\( -\log_{10}(6.6 \times 10^{-13}) = 12.18 \)[/tex]).
- Then we use [tex]\( \text{pH} = 14 - \text{pOH} \)[/tex].
- So, [tex]\( \text{pH} = 14 - 12.18 = 1.82 \)[/tex].
- Since pH [tex]\( 1.82 < 7 \)[/tex], the solution is Acidic.
To summarize:
- [tex]$\left[H^{+}\right] = 8.0 \times 10^{-2}$[/tex]: Acidic
- pH = 3.05: Acidic
- [tex]$\left[H^{+}\right] = 1.0 \times 10^{-7}$[/tex]: Neutral
- pH = 7.00: Neutral
- [tex]$\left[H^{+}\right] = 5.3 \times 10^{-8}$[/tex]: Basic
- [tex]$\left[OH^{-}\right] = 3.7 \times 10^{-5}$[/tex]: Basic
- pH = 8.92: Basic
- [tex]$\left[OH^{-}\right] = 6.6 \times 10^{-13}$[/tex]: Acidic
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