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What is the Kw of pure water at 50 degrees Celcius, if the pH is 6.630?

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Final answer:

To find the ionization constant of water (Kw) at 50 degrees Celsius with a pH of 6.630, we first calculate the concentration of hydronium ions from the pH, then square this value, as Kw equals the squared concentration of hydronium ions in pure water at equilibrium.

Step-by-step explanation:

The student is asking for the ionization constant, Kw, of pure water at 50 degrees Celsius given a pH of 6.630. In chemistry, the ionization constant is used to describe the equilibrium constant for the ionization of water, which varies with temperature.

To calculate Kw at 50°C, we first need to find the concentration of hydronium ions [H3O+], which can be obtained from the pH value:

[H3O+] = 10-pH = 10-6.630≈ 2.34 × 10-7M

Since the product of the concentrations of hydronium ions and hydroxide ions equals the ionization constant (Kw):

Kw = [H3O+][OH-]

For pure water, [H3O+] is equal to [OH-], thus:

Kw = (2.34 × 10-7M)2 = 5.48 × 10-14

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