Question

why water maintains PH under Auto-ionization?

Answer 1

Question #48299, Chemistry, Inorganic Chemistry

why water maintains PH under Auto-ionization?

Answer:

Autoionization is the "breaking apart" of water molecules to form H+ and OH-. For each water molecule you get a balance between "acid" and "base" making the water neutral. The concentrations of H+ (1.00x10-7) and OH- (1.00x10-7) are what give water its pH of 7.00 at 25°C.

At standard temperature and pressure (STP), the equilibrium constant of water, Kw, is equal to

$$\mathrm {K} _ {\mathrm {w}} = \left[ \mathrm {H} 3 \mathrm {O} ^ {+} \right] \left[ \mathrm {O H} ^ {-} \right]$$ $$K _ {w} = [ 1. 0 \times 1 0 ^ {- 7} ] [ 1. 0 \times 1 0 ^ {- 7} ]$$ $$\mathrm {K} _ {\mathrm {w}} = 1. 0 \times 1 0 ^ {- 1 4}$$

In this equation $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ is the concentration of hydronium ions, which in a chemical equation is the acid constant, Ka. The $\left[\mathrm{OH}^{-}\right]$ is the concentration of hydroxide ions, which in a chemical equation is the base constant, $K_{\mathrm{b}}$ . If given a pH, then you can easily calculate the $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ by simply taking the negative reverse log of the pH:

$$[ \mathrm {H} 3 \mathrm {O} + ] = 1 0 ^ {- \mathrm {p H}}.$$

The same formula applies to obtaining $\left[\mathrm{OH}^{-}\right]$ from the pOH:

$$[ \mathrm {O H} - ] = 1 0 ^ {- \mathrm {p O H}}$$

Adding the pH's gives you the $\mathsf{pK}_{\mathrm{w}}$

$$\mathrm {p K} _ {\mathrm {w}} = \mathrm {p H} + \mathrm {p O H} = 1 4. 0 0$$

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