Equilibrium concentration of a hydronium ion?
As it is expressed in your question, the reactions of weak acids and bases are equilibrium reactions, because they cannot ionize completely. Therefore it is better to represent such reactions with double arrows. HX(aq) + H2O(l) <---> H3O+(aq) + X-(aq) . Ka = 1.00x10^-06 Ka = [H3O+][X-] / [HX] Now we have to decide to which side the equilibrium will shift. Suppose 0.0530 M HX acid was in pure water, the dissociation would be; HX(aq) + H2O(l) <---> H3O+(aq) + X-(aq) (0.0530 - x) M ............... x M ............. x M Ka = [H3O+][X-] / [HX] 1.00x10^-06 = (x)(x) / (0.0530 -x) Now we make an assumption: Since x will be very small compared to 0.0530 it can be neglected. 1.00x10^-06 = (x)(x) / (0.0530) x^2 = 5.3 x 10^-8 x = 2.3 x 10^-4 M = [H3O+] = [X-] This also shows that our assumption is valid. Now let's consider our problem. In our problem [X-] = 0.207 M. This value is much more larger than [X-] present in pure water. This shows that the equilibrium will shift left. HX(aq) + H2O(l) <---> H3O+(aq) + X-(aq) (0.0530 + x) M <........... x M .......... 0.207 M Ka = [H3O+][X-] / [HX] 1.00x10^-06 = (x)(0.207) / (0.0530 +x) Again, the same assumption: x will be very small compared to 0.0530 , therefore it can be neglected. 1.00x10^-06 = (x)(0.207) / (0.0530) x = 3.9 x 10^-6 M = [H3O+]
What is the equilibrium concentration of hydronium ion in a solution?
HX is a weak acid that reacts with water according to the following equation: HX(aq) + H2O(l) H3O1+(aq) + X1-(aq) Ka = 1.58e-05 What is the equilibrium concentration of hydronium ion in a solution that is 0.0944 M in HX and 0.136 M in X1- ion? [H3O1+] = ____M
The equilibrium concentrations of H3O+, A-, and HA in a 0.04 M solution of the acid?
When you do these calculations for a very weak acid - especially with Ka = 4.0*10^-9, it is usual to consider that [HA] in the solution is equal to the molarity of the acid - in this case 0.04M. But as the question specifically asks for [HA] we will calculate it Let [H+] = X Then [A-] = X [HA] = (0.04-X) Equation: Ka = [H+] *[A-] / [HA] Substitute: 4.0*10^-9 = X² / (0.04-X) X² = (4.0*10^-9) * (0.04-X) X² = 1.6*10^-10 - ( 4.0*10^-9)X X² + (4.0*10^-9)X - (1.6*10^-10) = 0 Solve the quadratic for X X = 1.26*10^-5 that is the positive root. Answers: [H+] = [H3O+] = 1.26*10^-5 [A-] = 1.26*10^-5 [HA] = 0.04-(1.26*10^-5) = 0.039987 ( as I said, it is usual to assume that [HA] is the molarity of the acid. Doing this in the Ka equation avoids the complexity of a quadratic equation. )
Is the dynamic equilibrium concentration of reactants and products equal or constant?
Dynamic equilibrium means just that. it’s the dynamics that are equal, not the concentrations. The forward and the reverse rates of the process are equal, which means that the concentrations never change once the process reaches equilibrium. Some processes reach equilibrium very early in the reaction.For example, a weak acid, acetic acid, dissolved in water at a concentration of 1 mol/L will only react with water about 15% into the acetate ion and the hydronium ion. at that point, it has reached equilibrium and as fast at more acetic acid reacts with water to form acetate, the acetate is also reacting to go back to acetic acid. So nothing happens on the macroscopic scale even though the individual molecules are still reacting (the microscopic scale).On the other hand, a strong acid like HCl will react with water “nearly” to completion (i.e. 100%) to form [math]Cl^-[/math] and [math]H_3O^+[/math]. Note I quoted the word “nearly”. The Ka constant for [math]HCl[/math] in water is [math]1.3 \times 10^6.[/math] This large a value for Ka means that when the reaction reaches equilibrium, the concentration of [math]HCl[/math] is so small that we can ignore it and [[math]H_3O^+[/math]] is so close to what the original [[math]HCl[/math]] was that we just say it’s exact. In other words, even a strong acid or base will reach equilibrium but we assume they go 100% because most of the time, we are working to 2 or 3 sig figs of accuracy and our assumption of 100% works fine for that level of accuracy.
What is the difference between strong acids or bases and weak acids or bases?
Firstly, the definitions:Acids dissolve in water to form H+ ions.Bases dissolve in water to form OH- ions.Lets take a look at acids - A strong acid (HCl) vs a weak acid (CH3COOH).A strong acid will fully dissociate in water to form H+ ions.HCl + H2O---> H3O+ + Cl- This reaction is non-reversible. After dissolution, only a very very minute concentration of HCl itself remains in the solution, as most of the diluted HCl has dissolved into ions.Ka = [H+] [Cl-] / [HCl]A weak acid, however undergoes a reversible reaction as it only partially dissociates.CH3COOH + H2O <---> CH3COO- + H3O+Ka = [CH3COO-] [H3O+] / [CH3COOH]As Ka is a constant, as more CH3COOH is added, by Le Chatelier's Principle, in this reversible reaction, the system will aim to shift the equilibrium as such to reverse the change. Therefore, less CH3COOH than added will dissociate into its constituent ions.Ionic equilibrium can be explained by http://www.chem1.com/acad/webtex...The reason why one reaction is reversible and the other is not can be explained by http://chemed.chem.purdue.edu/ge...The favourability of a reaction (whether forward or reverse) will depend largely on the temperature and change in entropy.The same applied for bases. The only difference is that the base dissociates to form OH- ions instead.Strong and weak bases will depend likewise on whether the reaction is reversible (there are exceptions to this of course)NaOH + H2O ---> Na+ + OH- + H2OKb = [Na+] [OH-] / [NaOH]
Why is the increase in H30+ concentration slower than increase in weak acid concentration?
It is not a question of being slower, it is a question of equilibrium. I suppose that what you mean is why does the concentration of H3O+ ions increase as the concentration of the original acid for a strong acid, but not for a weak acid.A strong acid such as HCl is (to all practical purposes) fully ionized on water (99% or better), so one mole of HCl gives one mole of H3O+ ions. However, a weak acid such as acetic acid is only partially ionized and is in equilibrium, commonly represented asHOAc + H2O = H3O+ + OAc- where OAc means the acetate group.It depends on the individual acid what the degree of dissociation is, and there is an acid dissociation constant Ka quoted for individual substances. As water is usually in considerable excess, its concentration can be counted as constant. Therefore the equation can be given asHOAc = H+ + OAc-The equilibrium concentration quoted is therefore the ratio of[H+] [OAc-]/[HOAc] = KaGoing back to the strong acid, if we increase the concentration by a factor of 10, then the concentration of H+ will also go up by (about) the same factor.On the other hand for acetic acid, for the top to match the bottom, the value of [H+} and [OAc-] must increase as the square root of 10, or about 3.Since [H+] and [0Ac-] must be equalKa = constant = [H+]^2 / [HOAc]Thus we find the H+ concentration for 0.01 M acetic acid is about 0.0013, and for 0.1 M acetic is about 0.004.
Chemistry Problem - Equilibrium concentration of HONO?
The chemical equation would be: HONO + H2O --> H3O+ + ONO So the expression for Ka = [H3O+][ONO] / [HONO] The water is removed from the expression since it is always 56M so it is constant, since it is so large compared to everything else. Since the HONO is a weak acid, the equilibrium will lie to the left. This means that the change in [HONO] is negligible and so we treat [HONO] as constant. So for a 0.8 M solution, with a Ka of 5.1x10^-4 the expression becomes: 5.1 x 10^-4 = [H3O+][ONO-] / 0.8 Since [H3O+] = [ONO-] (1:1 molar ratio) we can rewrite it this way: 5.1 x 10^-4 = [H3O+]^2 / 0.8 Now solve for [H3O+]: [H3O+]^2 = 4.08 x 10^-4 [H3O+] = sqrt(4.08 x 10^-4) = 2.02 x 10^-2 M So the pH = -log(2.02 x 10^-2) = 1.69
How do chemists master applications of aqueous equilibria?
I choose to classify aqueous equilibria into two main categories:Acid-base equilibriaSolubility equilibriaFor 1, the main concepts here can be broken down and understood systematically using an Initial, Change, Eqm (ICE) table. Something like this:There are a few simplifying equations (assuming degree of ionization is <5%) below to calculate equilibrium concentration of [math]H_3O^{+}[/math] and [math]OH^-[/math] for a weak acid/base, where [math]C_o[/math] is the original concentration of weak acid/base:[[math]H_3O^{+}]=\sqrt{K_aC_o}[/math][[math]OH^-]=\sqrt{K_bC_o}[/math]pH = -lg [math]\sqrt{K_aC_o}[/math]pOH = -lg [math]\sqrt{K_bC_o}[/math]If the simplifying equations fail, return to using the good ol’ [math]K_a[/math] and [math]K_b[/math] equations.For buffer solutions, the solution must contain:For acidic buffers: Weak acid and its conjugate base.For basic buffers: Weak base and its conjugate acid.Do also remember the Henderson–Hasselbalch equation for calculating pH of buffer: [math]pH = pK_a + lg \frac {[conj. base]}{[acid]}[/math]Note: The same can be applied for basic buffers to calculate pOH.For 2, the main concepts are similar to acid-base equilibria, just that now, you are dealing with dissolving of ionic salts, and you don't use [math]K_a[/math] and [math]K_b[/math], but [math]K_{sp}[/math] and ionic product (or IP in short).You have to understand how the system changes when there is the presence of a common ion. For example, you are trying to dissolve [math]Mg(OH)_2[/math] in water but [math]NaOH[/math] is added. The presence of a common ion reduces the solubility of [math]Mg(OH)_2[/math] and this can be mapped out using the ICE table too.As long as IP>[math]K_{sp}[/math], the salt will be precipitated out, assuming no chemical reaction with the substance (can be anhydrous/in solution form) added in.Hope this helps!
Chemistry Homework Question: A solution 0.1 m of acetic acid (at 298 K) is partially dissociated in its ions. Ka=1.75*10^-5 mol/kg. What is the molality of the species in equilibrium if I add to the previous solution 0.1 moles of NO3K?
Aha! A trick Question!!!The trick is that potassium nitrate, NO3K, has no ions in common with acetic acid. It is neither an acid nor a base, so will not affect the hydronium ion concentration, so in fact, this substance can be treated as a separate non-interacting salt solution. It also dissociates completely.The second interesting thing about this question is that it is asking for molality, not molarity. Fortunately, the initial concentration and the Ka are all expressed in molality so we won't have to convert from molarity to molality.So let's start with the acetic acid. There are three soluble species possible, the undissociated acid (HOAC), the acetate anion (OAC-) and the hydronium ion, (H+). At equilibrium; HOAC <==> H+ + OAC-, Ka = 1.75 E-5 mol/kgInitial concs: 0.1m 1E-7 0.0mequilibrium: 0.1-x 1E-7+x xKa = [H+][OAC-]/[HOAC]1.75E-5 = [1E-7+x][x]/[0.1-x]The bad news is that you will have to solve this by iteration as x^2 is not significantly different from 1.76E-6*x, but the good news is that you will only need to do one iteration. Even better, x >> 1E-7, so the H+ concentration can ignore the initial H+.So with that and determining that x is 1.314E-3, we are ready to determine the concentrations of all the species:HOAC at equilibrium is (0.1-1.314E-3)H+ at equilibrium is 1.314E-3OAC- at equilibrium is 1.314E-3K+, as discussed above, assuming you started out with a kg of solvent, something that was not explicitly stated in the question, will be 0.1 mNO3- like the K+ will be 0.1 m.I hope this helped!