Answer:
A: [H⁺] = 2.4 × 10⁻¹¹ M
B: [H⁺] = 1.2 × 10⁻⁶ M
C: [H⁺] = 1.0 × 10⁻⁸ M
D: A and C are basic and B is acid.
Explanation:
Part A: Calculate [H⁺] for [OH⁻] = 4.1×10⁻⁴ M.
We know the ion-product of water Kw is:
Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]
Then,
![[H^{+} ]=\frac{K_{w}}{[OH^{-} ]} =\frac{1.0\times 10^{-14} }{4.1\times 10^{-4}} =2.4\times10^{-11} M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%3D%5Cfrac%7BK_%7Bw%7D%7D%7B%5BOH%5E%7B-%7D%20%5D%7D%20%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-14%7D%20%20%7D%7B4.1%5Ctimes%2010%5E%7B-4%7D%7D%20%3D2.4%5Ctimes10%5E%7B-11%7D%20M)
Part B: Calculate [H⁺] for [OH⁻] = 8.5×10⁻⁹ M.
Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]
![[H^{+} ]=\frac{K_{w}}{[OH^{-} ]} =\frac{1.0\times 10^{-14} }{8.5\times 10^{-9}} =1.2\times10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%3D%5Cfrac%7BK_%7Bw%7D%7D%7B%5BOH%5E%7B-%7D%20%5D%7D%20%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-14%7D%20%20%7D%7B8.5%5Ctimes%2010%5E%7B-9%7D%7D%20%3D1.2%5Ctimes10%5E%7B-6%7D%20M)
Part C: Calculate [H⁺] for a solution in which [OH⁻] is 100 times greater than [H⁺].
We know that [OH⁻] = 100 . [H⁺]. If we replace this in the ion-product of water:
Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]
1.0 × 10⁻¹⁴ = [H⁺]. 100. [H⁺]
1.0 × 10⁻¹⁴ = 100 . [H⁺]²
1.0 × 10⁻¹⁶ = [H⁺]²
[H⁺] = √1.0 × 10⁻¹⁶ = 1.0 × 10⁻⁸ M
Part D: Indicate whether the solution is acidic, basic, or neutral.
A solution is acid when [H⁺] > 10⁻⁷ M, basic when [H⁺] < 10⁻⁷ M and neutral when [H⁺] = 10⁻⁷M. So A and C are basic and B is acid.