To solve this problem, we need to understand what pH values represent and how they relate to the hydronium ion ([H₃O⁺]) concentration in a solution.
pH is defined as the negative logarithm (base 10) of the hydronium ion concentration:
[tex]\[ \text{pH} = -\log[\text{H₃O⁺}] \][/tex]
Given:
- Initial pH = 6.0
- Final pH = 4.0
Step-by-step:
1. Calculate the initial hydronium ion concentration:
[tex]\[ \text{Initial [H₃O⁺]} = 10^{-\text{pH}_{\text{initial}}} \][/tex]
[tex]\[ \text{Initial [H₃O⁺]} = 10^{-6.0} \][/tex]
[tex]\[ \text{Initial [H₃O⁺]} = 1 \times 10^{-6} \, \text{M} \][/tex]
2. Calculate the final hydronium ion concentration:
[tex]\[ \text{Final [H₃O⁺]} = 10^{-\text{pH}_{\text{final}}} \][/tex]
[tex]\[ \text{Final [H₃O⁺]} = 10^{-4.0} \][/tex]
[tex]\[ \text{Final [H₃O⁺]} = 1 \times 10^{-4} \, \text{M} \][/tex]
3. Determine the factor of change in hydronium ion concentration:
To find how much the hydronium ion concentration has changed, we take the ratio of the final concentration to the initial concentration:
[tex]\[ \text{Factor Change} = \frac{\text{Final [H₃O⁺]}}{\text{Initial [H₃O⁺]}} \][/tex]
[tex]\[ \text{Factor Change} = \frac{10^{-4}}{10^{-6}} \][/tex]
[tex]\[ \text{Factor Change} = 10^{-4} \times 10^{6} \][/tex]
[tex]\[ \text{Factor Change} = 10^{2} \][/tex]
[tex]\[ \text{Factor Change} = 100 \][/tex]
Since the concentration increased by a factor of 100, the correct phrase to describe the change in the hydronium ion concentration of the solution is:
(4) increased by a factor of 100
