4. To calculate the [H+] and [NO3-] of a 0.175M solution of nitric acid, we need to consider that nitric acid is a strong acid and fully dissociates in water. Therefore, for every 1 mol of nitric acid, we will get 1 mol of H+ and 1 mol of NO3-.
Therefore, the [H+] and [NO3-] in the 0.175M solution of nitric acid will also be 0.175M.
[H+] = 0.175 M
[NO3-] = 0.175 M
5. Since HCl is also a strong acid, it fully dissociates in water. Therefore, the concentration of hydroxide ions (OH-) in a solution of HCl will be negligible.
6. According to the neutralization reaction between Ca(OH)2 and water, 1 mol of Ca(OH)2 produces 2 moles of OH-. Therefore, the concentration of OH- in a 0.01M solution of Ca(OH)2 will be double the concentration of Ca(OH)2.
[OH-] = 2 * 0.01 M = 0.02 M
7. The ionization constant of water, Kw, at 37Β°C is 1.00Γ10^-13.60. At equilibrium, the concentration of H3O+ (hydroxonium ions) will be equal to the concentration of OH-. Therefore, we can take the square root of Kw to calculate the concentrations of H3O+ and OH-.
[H3O+] = [OH-] = sqrt(Kw)
= sqrt(1.00Γ10^-13.60)
8.
(a) The pH of a solution can be calculated using the equation pH = -log[H+]. For 0.0028M HCl, the concentration of H+ is 0.0028M.
pH = -log(0.0028)
(b) For 0.014M HNO3, the concentration of H+ is 0.014M.
pH = -log(0.014)
(c) For 0.00052M HBr, the concentration of H+ is 0.00052M.
pH = -log(0.00052)
(d) For 0.092M HI, the concentration of H+ is 0.092M.
pH = -log(0.092)
9. The ionization constant of ammonia, Kb, is given as 1.75Γ10^-5. Since ammonia is a weak base, it does not completely dissociate in water. To calculate the pH of a 0.025M solution of ammonia, we need to consider the equilibrium of the reaction between ammonia and water: NH3 + H2O β NH4+ + OH-.
[H+] can be calculated using the Kb value:
Kb = [NH4+][OH-] / [NH3]
[H+] = sqrt(Kw * Kb * [NH3])
where [NH3] is the concentration of ammonia, which is 0.025M.
Then, the pH can be calculated as pH = -log[H+].