The equilibrium concentration of OH⁻ at this temperature is 6.708 × 10⁻⁸ M
Ionic product of water
The ionic product of water Kw = [H⁺][OH⁻] where
- [H⁺] = hydrogen ion concentration and
- [OH⁻] = hydroxide ion concentration.
At equilibrium, [H⁺] = [OH⁻]
So, [H⁺][OH⁻] = Kw
[OH⁻][OH⁻] = Kw
{OH⁻}² = Kw
Taking square root of both sides, we have
√{OH⁻}² = √Kw
Equilibrium concentration of OH⁻
{OH⁻} = √Kw
Since Kw at 15 °C is 4. 5 × 10⁻¹⁵, substituting the value of the variable into the equation, we have
{OH⁻} = √Kw
{OH⁻} = √(4. 5 × 10⁻¹⁵)
{OH⁻} = √(45 × 10⁻¹⁶)
{OH⁻} = √45 × √10⁻¹⁶
{OH⁻} = 6.708 × 10⁻⁸ M
So, the equilibrium concentration of OH⁻ at this temperature is 6.708 × 10⁻⁸ M
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