Calculate the concentration of [H+] and [OH-] in 0.002 mol/L of HNO3 solution.
Please answer
To calculate the concentration of [H+] and [OH-] in a solution of HNO3, we need to consider the dissociation of the acid.
The balanced equation for the dissociation of HNO3 in water is:
HNO3(aq) → H+(aq) + NO3-(aq)
Since HNO3 is a strong acid, it dissociates completely in water, which means that one mole of HNO3 will produce one mole of H+ ions.
Therefore, the concentration of [H+] in a 0.002 mol/L HNO3 solution is also 0.002 mol/L.
Since water is a neutral substance, the concentration of [OH-] in a HNO3 solution is negligible compared to the concentration of [H+]. Therefore, we can assume that the [OH-] concentration is practically zero.
To calculate the concentration of [H+] and [OH-] in a solution of HNO3, we need to understand the properties of the acid and the relationship between [H+] and [OH-] in water.
HNO3 is a strong acid, which fully dissociates in water to form H+ and NO3- ions. Since HNO3 is a monoprotic acid, each molecule of HNO3 contributes one H+ ion to the solution.
The concentration of [H+] in the HNO3 solution is therefore equal to the concentration of the HNO3 itself, which is given as 0.002 mol/L.
In pure water, at 25°C, the concentrations of [H+] and [OH-] are related by the equation: [H+][OH-] = 1.0 x 10^-14 mol^2/L^2. This is known as the ion product of water.
Since HNO3 is a strong acid and fully dissociates, we can assume that the concentration of [H+] is much greater than the concentration of [OH-] in the solution. Therefore, we can neglect the contribution of [OH-] to the ion product of water calculation.
Thus, we can calculate the concentration of [OH-] using the ion product of water equation:
[H+][OH-] = 1.0 x 10^-14 mol^2/L^2
[OH-] = (1.0 x 10^-14 mol^2/L^2) / [H+]
Plug in the value of [H+] as 0.002 mol/L:
[OH-] = (1.0 x 10^-14 mol^2/L^2) / (0.002 mol/L)
Simplifying the equation, we find:
[OH-] = 5.0 x 10^-12 mol/L
Therefore, the concentration of [H+] in the HNO3 solution is 0.002 mol/L, and the concentration of [OH-] is 5.0 x 10^-12 mol/L.
The H^+ in 0.002 M HNO3 is 0.002. There is some H^+ from the ionization of H2O but it is small enough to be ignore. For the OH, use
HOH ==> H^+ + OH^-
Kw = 1E-14 = (H^+)(OH^-)
You know Kw and H^+ (from the HNO3), solve for OH^- in the system.