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Sagot :
To determine the hydronium ion concentration in a solution of HCl with a given pH value and the hydroxide ion concentration in a solution of NH₃ with a given pOH value, follow these steps:
### Step 1: Calculate the hydronium ion concentration [H₃O⁺]
Given:
- pH = 4.65
The relationship between pH and hydronium ion concentration is given by:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \][/tex]
Substitute the given pH value:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-4.65} \][/tex]
Next, we determine the value:
[tex]\[ [\text{H}_3\text{O}^+] \approx 0.0 \][/tex]
Therefore, the hydronium ion concentration, rounded to the nearest hundredth, is:
[tex]\[ [\text{H}_3\text{O}^+] = 0.0 \times 10^{-4} \, M \][/tex]
And the exponent [tex]\( n \)[/tex] is:
[tex]\[ n = -4 \][/tex]
### Step 2: Calculate the hydroxide ion concentration [OH⁻]
Given:
- pOH = 4.65
The relationship between pOH and hydroxide ion concentration is given by:
[tex]\[ [\text{OH}^-] = 10^{-\text{pOH}} \][/tex]
Substitute the given pOH value:
[tex]\[ [\text{OH}^-] = 10^{-4.65} \][/tex]
Next, we determine the value:
[tex]\[ [\text{OH}^-] \approx 0.0 \][/tex]
Therefore, the hydroxide ion concentration, rounded to the nearest hundredth, is:
[tex]\[ [\text{OH}^-] = 0.0 \times 10^{-4} \, M \][/tex]
And the exponent [tex]\( n \)[/tex] is:
[tex]\[ n = -4 \][/tex]
### Summary
- Hydronium ion concentration in HCl with pH of 4.65:
[tex]\[ 0.0 \times 10^{-4} \, \text{M} \\ n = -4 \][/tex]
- Hydroxide ion concentration in NH₃ with pOH of 4.65:
[tex]\[ 0.0 \times 10^{-4} \, \text{M} \\ n = -4 \][/tex]
This concludes our calculations and detailed, step-by-step solution.
### Step 1: Calculate the hydronium ion concentration [H₃O⁺]
Given:
- pH = 4.65
The relationship between pH and hydronium ion concentration is given by:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \][/tex]
Substitute the given pH value:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-4.65} \][/tex]
Next, we determine the value:
[tex]\[ [\text{H}_3\text{O}^+] \approx 0.0 \][/tex]
Therefore, the hydronium ion concentration, rounded to the nearest hundredth, is:
[tex]\[ [\text{H}_3\text{O}^+] = 0.0 \times 10^{-4} \, M \][/tex]
And the exponent [tex]\( n \)[/tex] is:
[tex]\[ n = -4 \][/tex]
### Step 2: Calculate the hydroxide ion concentration [OH⁻]
Given:
- pOH = 4.65
The relationship between pOH and hydroxide ion concentration is given by:
[tex]\[ [\text{OH}^-] = 10^{-\text{pOH}} \][/tex]
Substitute the given pOH value:
[tex]\[ [\text{OH}^-] = 10^{-4.65} \][/tex]
Next, we determine the value:
[tex]\[ [\text{OH}^-] \approx 0.0 \][/tex]
Therefore, the hydroxide ion concentration, rounded to the nearest hundredth, is:
[tex]\[ [\text{OH}^-] = 0.0 \times 10^{-4} \, M \][/tex]
And the exponent [tex]\( n \)[/tex] is:
[tex]\[ n = -4 \][/tex]
### Summary
- Hydronium ion concentration in HCl with pH of 4.65:
[tex]\[ 0.0 \times 10^{-4} \, \text{M} \\ n = -4 \][/tex]
- Hydroxide ion concentration in NH₃ with pOH of 4.65:
[tex]\[ 0.0 \times 10^{-4} \, \text{M} \\ n = -4 \][/tex]
This concludes our calculations and detailed, step-by-step solution.
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