The value of [H₃O⁺] and pH of water at 0°C and 50°C.
0°C value of [H₃O⁺] = 3.375 x 10⁻⁸
50°C value of [H₃O⁺] = 2.340 x 10⁻⁷
pH of water at 0°C = pH = 7.4717
pH of water at 50°C = pH = 6.6308
<h3>pH of water at different level:</h3>
Water with a pH less than 7 is considered acidic, while water with a pH greater than 7 is considered basic. The normal pH range for surface water systems is 6.5 to 8.5, and for groundwater systems is 6 to 8.5. Alkalinity is a measure of a water's ability to withstand a pH change that would cause it to become more acidic.
<h3>According to the given information:</h3>
0°C = Kw = 1.139 × 10⁻¹⁵
50°C = 5.474 × 10⁻¹⁴
Solving at 0°C for [H₃O⁺] and pH. water is neutral therefore its[H₃O⁺] [OH⁻]
are equal [H₃O⁺] = [OH⁻].
Kw = [H₃O⁺] [OH⁻]
Kw = [H₃O⁺]²
[H₃O⁺] = √Kw
= √1.139 × 10⁻¹⁵
= 3.375 x 10⁻⁸
pH = -log[H₃O⁺]
= -log 3.375 x 10⁻⁸
= 7.4717
Solving at 50°C for [H₃O⁺] and pH. water is neutral therefore its[H₃O⁺] [OH⁻]
are equal [H₃O⁺] = [OH⁻].
Kw = [H₃O⁺] [OH⁻]
Kw = [H₃O⁺]²
[H₃O⁺] = √Kw
= √ 5.474 × 10⁻¹⁴
= 2.340 x 10⁻⁷ M
pH = -log[H₃O⁺]
= -log2.340 x 10⁻⁷
pH = 6.6308
The value of [H₃O⁺] and pH of water at 0°C and 50°C.
0°C value of [H₃O⁺] = 3.375 x 10⁻⁸
50°C value of [H₃O⁺] = 2.340 x 10⁻⁷
pH of water at 0°C = pH = 7.4717
pH of water at 50°C = pH = 6.6308
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I understand that the question you are looking for is:
Kw = 1.139 × 10⁻¹⁵ at 0°C and 5.474 × 10⁻¹⁴, find [H₃O⁺] and pH of water at 0°C and 50°C. find the [H₃O⁺] and pH of water at 0°C and 50°C.
