3. Consider the following equilibrium for nitrous acid, HNO2, a weak acid:?
(a) right as a competing equilibrium uses up much of the H3O+(aq) so forward reaction exceeds reverse reaction until equilibrium re-established (consult Le Chatelier's Principle) (b) left as NO2-(aq); a product of the equilibrium, is added.... (c) left as H3O+(aq) added as a constituent of the HCl solution; and if HCl gas is used instead, it dissolves in the water to form H3O+(aq) as well. The amount of H2O consumed is such that it still stays in huge excess anyway (d) one assumes this is done by adding water, so the concentrations of all ions is decreased, and reaction will proceed to the right to increase the number of particles in solution back towards a level that gets more particles/ions.... The last one is best described mathematically. If we double the volume of water, [H2O] is basically unchanged - still over 55M whereas the concentration of the remaining three partners in the equilibrium is instantly halved. Using the equilibrium law: K...=..[H3O+]..x..[NO2-] .........._____________ ..........[HNO2]..x..[H2O] becomes 0.5 x 0.5 .............._______ .............0.5 x constant which is a fraction that is a half of the equilibrium constant (0.25/0.5 = 0.5) the top line values have to increase and the bottom line ones decrease to get back to unity (the equilibrium constant) Product concentrations have to increase or reactant concentrations have to decrease, or both. Both it is, so reaction shifts to the right to re-establish equilibrium.
Consider the following reaction: HC2H3O2(aq)+H2O(l)⇌H3O+(aq)+C2... Kc=1.8×10−5 at 25∘C?
HC2H3O2 + H2O <====> H3O+ + C2H3O- K = [H3O+] [C2H3O2-] / [HC2H3O2] Let x = change in [HC2H3O2] At equilibrium: [HC2H3O2] = (0.250 – x) At equilibrium = [H3O+] = x and [C2H3O2-] = x 1.8 x 10^-5 = (x) (x) / (0.250 – x) ; I will neglect x in the (0.250 –x) since x is much smaller than 0.250 x^2 = 1.8 x 10^-5(0.250) x^2 = 4.6 x 10^-6 x = [H3O+] = 2.1 x 10^-3
Consider the following reaction (equilibrium concentration)?
The other answer started correctly, but had an error. So, the expression for Kc is: Kc = 1.8 X 10^-5 = [H3O+][C2H3O2-]/[HC2H3O2] Let x be mol/L of H3O formed. You will also form x mol/L of C2H3O2- at the same time, and you will use up x mol/L of HC2H3O2. So, Kc = 1.8 X 10^-5 = x*x / (0.190 - x) assume x<<0.190. So, 1.8 X 10^-5 = x^2 / 0.190 x^2 = 3.4X10^-6 x = [H3O+] = 1.85 X 10^-3 mol/L
Consider the reaction HC2H3O2(aq) +H2O(l)-->H3O+(aq)+C2H3O2 to the negative power(aq).which species is the ?
H3O+. H2O acts as a base on the left side of the equation, and its pair is H3O+. If the reaction were reversed the H3O+ would be the proton donor, making it the conjugate acid.
Will HCl or H2SO4 (same concentrations in molarity) require more volume (in mL) to neutralize a solution of NaOH?
All of the other answers here are quite correct, however I thought I’d expand a little bit on the reason why sulphuric acid neutralises twice the molar quantity of NaOH that HCl does. Robert Goodman has hinted at the reason in his answer where he mentions Normality versus Molarity.The reaction of sulphuric acid and sodium hydroxide is normally written as;-H2SO4 + 2 NaOH —→ Na2SO4 + 2H2OFrom this it is clear that one mole of sulphuric acid will neutralise two moles of sodium hydroxide. However the reaction, as written above, is actually the overall reaction resulting from several separate, intermediate reactions.Sulphuric acid is a diprotic acid, which means its ionisation in aqueous solution to produce protons occurs in a two separate stages with the intermediate formation of a bisulphate (or hydrogen sulphate) ion;-H2SO4(l) + H2O(l) —→ H3O+(aq) + HSO4-(aq)HSO4-(aq) + H2O(l) —→ H3O+(aq) + SO4–2(aq)Each reaction releases one mole of hydronium ions (H3O+), each of which can neutralise one mole of NaOH;-H3O+(aq) + OH- (aq) ——-> 2 H2O(l)Similarly, a triprotic acid, such as phosphoric acid (H3PO4), would neutralise three molar equivalents of sodium hydroxide… and so on.
Consider the following unbalanced equation for the neutralization of acetic acid.?
HC2H3O2(aq) + Ca(OH)2(aq)--> H2O(l) + Ca (C2H3O2)2(aq) 1.Balance the equation. Express your answer as a chemical equation. Identify all of the phases in your answer. 2. Determine how many moles of Ca (OH)2 are required to completely neutralize 3.23 mol of HC2H3O2. Thanks!
Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant?
HC2H3O2(aq) + H2O(l) <-----> H3O+(aq) + C2H3O2-(aq) 0.18 M - x ---> ---> ---> --> X & X but when K's are small like e-5's or smaller, we disregard subrtacting the X form the 0.18 since this is only a 2 sig fig problen=m & correcting 0.18 by subtracting the X is far beyond 2 sig figs Ka = [H3O+] [C2H3O2-] [HC2H3O2] 1.8 e-5 = [X] [X] / [0.18] X2 = 3.24 e-6 X = [H+] = 1.8 e-3 pH = 2.74