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$A\xrightleftharpoons[k_{-1}]{k_1}B\xrightleftharpoons[k_{-2}]{k_2}C\xrightleftharpoons[k_{-3}]{k_3}D$ I believe the rate of the overall reaction is $\frac{d[D]}{dt}$ = $k_3[C]$ since D is removed, $k_{-3}$ won't affect the reaction. Now we can apply steady-state approximation to B and C: Steady-state approximation for B: $\frac{d[B]}{dt} = 0 = -k_2[B] - k_{-... 2 Question: Determine the$\mathrm{pH}$of the solution resulting when$\pu{100 cm^3}$of$\pu{0.50 mol dm-3}\ce{CH2ClCOOH}$is mixed with$\pu{200 cm^3}$of$\pu{0.10 mol dm-3}\ce{NaOH}$. I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of ... 1$\ce{CH2ClCOOH}$is a weak acid;$\ce{NaOH}$is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid? They react. $$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O}$$ A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base. $$N_1V_1=N_2V_2$$ At this stage, an equivalence ... 0 The mistake in your approach is that you are assuming moles and activity to be same. Activities of solids and pure liquids are constant and thus we don't consider them while writing$Q$or$K$. They are already considered when defining the value of$K$. For example, we don't consider activity of water in the$K_w$as that's how it is defined. Now coming to ... 1 I know you wanted something non-mathematical, but I think the best way to understand this intutively requires some simple math. Suppose you have the following elementary reaction: $$\ce{A(g) + B(g) <=> C(g)}$$ By "elementary reaction" I mean that this shows the actual reaction mechanism, such that the rate equation can be obtained directly ... 2 This was answered in the comments, but these second equilibrium expression should not actually be a$K_b$value. Rather, it should be$1/K_a = 55600$. Using this value in the same calculations, we find that$x=5.5805\times 10^{-4}$and$y = 5.5795\times 10^{-4}$, so that$\ce{pH} = -\log(x-y) = 7.00$. 1 Why should they ( supposing ideal gas behaviour ) ? If a volume, containing$\pu{4 mol}\ce{N2}$and$\pu{1 mol}\ce{O2}$has pressure$\pu{5 atm}$, partial pressures are:$\ce{N2} : \pu{4 atm}\ce{O2} : \pu{1 atm}$If you add$\pu{5 mol}\ce{Ar}$, the total pressure raises to$\pu{10 atm}$, with partial pressures :$\ce{Ar} : \pu{5 atm}\ce{N2} : ...

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The free energy difference between reactants and products are given as follows when both are in their STANDARD states: $$\Delta G^\circ = -RT \ln K_\mathrm{eq}$$ If $K_\mathrm{eq}$ is less than one that means $\Delta G^\circ$ is positive and the reaction as written will go to the left until the equilibrium conditions are satisfied. At this time $\Delta G = 0$...

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The $\ce{Na+}$ ion is a spectator ion in this case. It doesn't participate in the chemical reaction. It is found unchanged on both the reagent and product sides (red top-right and green bottom-right reaction formulas), so it can be canceled out to get the net reaction (red left). From there, he did the usual acid-base calculation to get a formula for $\ce{Kb}... 0 Sodium ion is the weak conjugate acid of strong base i.e. NaOH. So the show which one ot of the two ions furnished undergoes hydrolysis to a greater extent the Na+(which does not react) is crossed and hydrolysis of stronger base C2H3O2- to produce OH- ions is shown. 0 The OP asked the question again, and proposed the following solution from his book. My textbook says to solve it in the following way: 1) Ignore the 0.10 M KI since it doesn't matter. 2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its 𝐾sp. 𝐾sp=1.77×10−10=𝑥^2 𝑥=1.3×10^−5 This is the Ag+ concentration 3) Substitute ... 1 You must first understand what happens when water is heated along the boiling curve. Start from the point$\pu{100°C}$and$\pu{1 atm}$. At this point there are two phases, one liquid whose density is nearly$\pu{1 g/cm^3}$and exactly$\pu{0.96 g/cm3}$. And the vapor has a density of$\ce{0.0006 g/cm^3}$. If now you heat this system in a closed volume, the ... 5 The essential condition is the compound thermal stability. If it decomposes below its melting point, it does not have a triple point. If it decomposes before properties of gaseous and liquid phases converge to each other, it has just estimated, extrapolated triple point. Beyond CP we talk about supercritical fluid, it kind of shares many properties of both ... 0 Acids are hydrogen ion donors. When the react with water, they can give a hydrogen ion to form$ \ce{H3O+}$. For example:$ \ce{HCl(aq) + H2O(l) \rightarrow Cl-(aq) + H3O+(aq)}$Simple acids, such as$ \ce{HCl}$or$\ce{H2SO4}$, can be recognized as acids by the H at the start of the formula. Other more complex acids may be written with$ \ce{COOH}$at ... 1 H3O+ is just the combination of a a$\ce{H+}$ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the$\ce{H+}$is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative ... 3 I will continue with the data of$[\ce{H2CO3}] = \pu{10^{-4.97} M}.$Now, as the$K_2$of$\ce{H2CO3}$is very small as compared to its$K_1,$we can assume that all the$\ce{H+}$will come from the first dissociation of$\ce{H2CO3}.$$$\begin{array}{lccc} & \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\ &\text{Initial} & ... 2 I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it. Given the reaction:$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}$$there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is ... 0 Since your weak acid is neutralized to the end point by a strong base, \ce{NaOH}, the final reding does not depend on the percentage of dissociation of the weak acid before neutralization. Suppose your ionization of the weak acid is as follows:$$\ce{HA + H2O <=> H3O+ + A-} \tag1$$Thus at the end of neutralization:$$\text{amount of$\ce{NaOH}$... 1 As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage. Is$\pu{[}\ce{HA}{] = 10^{−3} M}$or is$\pu{[}\ce{H+}{] = 10^{−3}}$?$\pu{[}\ce{HA}{] = 10^{-3}}$. Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base: $$N_1V_1=N_2V_2$$ Since$V_1=V_2$, ... 0 Assuming the water is at the same temperature and pressure on both sides, the only way to create a difference in pH would be through the presence of conjugate acids and/or bases other than$\ce{H_2O}$,$\ce{H^+}$and$\ce{OH^-}$. I.e., you can't have pure water at the same temperature and pressure on both sides, yet have a difference in pH. [Pressure would ... 0 While learning about acidic and basic buffer solutions we were told that they can resist pH change if we add small amounts of acid or base by neutralizing the$\ce{H+/OH−}$ions from the added acid or base. Yes, the pH changes less when you add small amounts of acid or base to a buffer, and more if you add it to pure water or an unbuffered aqueous solution. ... 1 Matrices, as shown in DHMO's answer, provide a way to solve systems of linear equations. But I think that, for most students, it might be more accessible to just solve the simultaneous linear equations directly. Specifically: We know that, in a balanced chemical equation, the number of atoms of each element on the left must equal the number of atoms of ... 2 I think your teacher's version of hydrolysis of sodium peroxide is straight out from Wikipedia article, which follows the reference 1. Wikipedia states that: Sodium peroxide hydrolyzes to give sodium hydroxide and hydrogen peroxide according to the reaction: $$\ce{Na2O2 + 2H2O -> 2NaOH + H2O2}$$ However, even though it can seemingly be considered to go ... 5$\ce{HO2-}$is a weaker base than$\ce{NaOH}$. but$\ce{O2^2-}$is much stronger base than$\ce{HO2-}\$ and does not occur in water solutions in significant amount. But its salts ( sometimes in the form of hydrates ) can be precipitated at highly alkaline solutions of hydrogen peroxide. Additionally, lack of product presence supports the respective ...

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