Answer:
2.90 x 10⁻¹¹moldm⁻³
Explanation:
Given parameters:
[H⁺] = 3.5 x 10⁻⁴mol/dm³
Unknown
[OH⁻] = ?
Solution;
The ionic product of water can be used to solve this problem. It has been experimentally determined to be 1 x 10⁻¹⁴mol² dm⁻⁶
[H⁺] [OH⁻] = 1 x 10⁻¹⁴
Therefore;
[OH⁻] = [tex]\frac{1 x 10^{-14} }{H}[/tex] = [tex]\frac{1 x 10^{-14} }{3.5 x 10^{-4} }[/tex] = 0.29 x 10⁻¹⁰moldm⁻³
= 2.90 x 10⁻¹¹moldm⁻³
