In a solution of pure water, the dissociation of water can be expressed by the following: H₂O(l) + H₂O(l) H₂O'(aq) + OH(aq) The

Question

In a solution of pure water, the dissociation of water can be
expressed by the following:
H₂O(l) + H₂O(l) H₂O'(aq) + OH(aq)
The equilibrium constant for the ionization of water, K, is called
the ion-product of water. In pure water at 25 °C, K has a value of
1.0 10¹. The dissociation of water gives one H₂O' ion and one OH
ion and thus their concentrations are equal. The concentration of
each is 1.0 x 10 M.
K - [HO][OH^]
K(1.0 10 )(1.0 × 10) - 1.0 × 10 ¹¹
X
X
[H₂O'][OH]- 1.0 × 10 ¹
X
A solution has a [OH ]-3.4 10 M at 25 °C. What is the [HO] of
the solution?
O
O
2.9 × 109 M
X
3.4 × 10⁹ M
2.9 × 10-¹5 M
2.9 × 10 10 M
I DON'T KNOW YET

Kunduz · Accepted Answer

Click to see the answer