Question

at 60 degrees celcius, the value of kw is 1.0 x 10 -13 exponent. calculate the concentrations of hydronium ions and hydroxide ions in water at this temperature.

Answer 1

at 60 degrees celcius, the value of kw is $1.0 \times 10^{-13}$ exponent. Calculate the concentrations of hydronium ions and hydroxide ions in water at this temperature.

Solution:

The equilibrium self-ionization of water is shown below:

$$2 \mathrm{H}_{2} \mathrm{O} \leftrightarrows \mathrm{H}_{3} \mathrm{O}^{+} + \mathrm{OH}^{-}$$

The product of hydronium and hydroxide ions concentration is the ion product of water:

$$\mathrm{K}_{\mathrm{w}} = \left[ \mathrm{H}_{3} \mathrm{O}^{+} \right] \cdot \left[ \mathrm{OH}^{-} \right]$$

So, the concentrations of $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ and $\left[\mathrm{OH}^{-}\right]$ calculate for the equation:

$$\left[ \mathrm{H}_{3} \mathrm{O}^{+} \right] = \left[ \mathrm{OH}^{-} \right] = \sqrt{\mathrm{K}_{\mathrm{w}}} = \sqrt{10 \cdot 10^{-13}} = 3.16 \cdot 10^{-7} \ \mathrm{mol/L}$$

Answer: $\left[\mathrm{H}_{3} \mathrm{O}^{+} \right] = \left[\mathrm{OH}^{-}\right] = 3.16 \cdot 10^{-7} \ \mathrm{mol/L}$