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Sulfuric acid, H2SO4,readily dissociates intoH+andHSO4-H+andHSO4-ions

H2SO4H++HSO4-

The hydrogen sulfate ion, in turn, can dissociate again:

HSO4-H++SO42-

The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 10 and 10*, respectively. (For dissociation of acids it is usually more convenient to look up K than G°. By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = -2, while for the second reaction pK = 1.9.)

(a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?

(b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically 5 x 10 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of SOg. What is the pH of this rainwater?

(c) Explain why you can neglect dissociation of water into H* and OH in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO and SO2-?

Short Answer

Expert verified

Therefore,

(a)pH=1(b)MHSO4-=3.9716×10-7mol/kg,pH=4(c)MOH-=1.0×10-10mol/kg(d)pH=1.9

Step by step solution

01

Given information

(a) We have two equations:

H2SO4H++HSO4-(1)HSO4-H++SO42-(2)

The equilibrium constant for the first reaction is:

role="math" localid="1647214128324" K=mH+×mHSO4-mH2SO4sinceKis large we can replacemH+withmHSO4-, so:K=mHSO4-2mH2SO4

The equilibrium constant is K=100

mHSO4-=10mH2SO4

When the morality of H2SO4 equals 1, then the morality of HSO4-, is 10, which implies that the morality of H+ is 10. Then the pH of the solution is:

pH=-log10mH+=-log10(10)=1

This indicates that the solution is highly acidic; the first equation's left side is acidic, implying that the reaction will move to the right. As a result, all of the sulfuric acid will be separated.

02

Explanation

(b)The equilibrium constant for first reaction is:

K=MH+×MSO42-MHSO4-

The equilibrium constant can be written in terms of the pK of the reaction, as:

pK=-log10(K)K=10-pK

Therefore,

10-pK=MH+×MSO42-MHSO4-

Each molecule of sulfuric acid gives two molecules of HT, thus the molality of the HT is twice the molality of the sulfuric acid, therefore we can write:

10-pK=2MSO42-×MSO42-MHSO4-10-pK=2MSO42-2MHSO4-MHSO4-=2MSO42-210-pK

If the concentration of the sulfate in rain water is MSO42-=5.0×10-5mol/kg, substitute this value

MHSO4-=25.0×10-5mol/kg210-1.9MHSO4-=3.9716×10-7mol/kg

The concentration of H+is

MH+=2MSO4-2=25.0×10-5mol/kg=10×10-5mol/kg

pH of rainwater is

pH=-log10MH+=-log1010×10-5mol/kg=4

03

Explanation

(c)The concentrations of the OH- and H+

MH+·MOH-=10-14MOH-=10-14MH+

Substitute the value of MH+

role="math" localid="1647244030613" MOH-=10-1410×10-5mol/kg=1.0×10-10mol/kg

04

Explanation

(d)When the dissolved sulfate is equally distrbuted between HSO4-andSO4-2we can set MSO42-equals to MHSO4-

role="math" localid="1647244228696" 10-pK=MH+×MSO42-MHSO4--MH+=10-pKMH+=10-1.9

Hence, pH of the solution is:

pH=-log10MH+=-log1010-1.9mol/kgpH=1.9

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Most popular questions from this chapter

By subtracting μNfrom localid="1648229964064" U,H,F,orG,one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),

ΦU-TS-μN.

(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives ofΦwith respect toT,V, and μN

(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φtends to decrease.

(c) Prove thatϕ=-PV.

(d) As a simple application, let the system be a single proton, which can be "occupied" either by a single electron (making a hydrogen atom, with energy -13.6eV) or by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800Kand an electron concentration of about 2×1019per cubic meter. Calculate Φfor both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in Problem 5.20, the prediction for such a small system is only a probabilistic one.)

Compare expression 5.68 for the Gibbs free energy of a dilute solution to expression 5.61 for the Gibbs free energy of an ideal mixture. Under what circumstances should these two expressions agree? Show that they do agree under these circumstances, and identify the function f(T, P) in this case.

An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at 1°C. The system is then placed in thermal contact with a low-temperature reservoir at -1°C until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at 1°C. The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor's reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, 1 -Te/Th

Figure 5.35 (left) shows the free energy curves at one particular temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition. Draw tangent lines to determine which phases are present at which values of x. To determine qualitatively what happens at other temperatures, you can simply shift the liquid free energy curve up or down (since the entropy of the liquid is larger than that of any solid). Do so, and construct a qualitative phase diagram for this system. You should find two eutectic points. Examples of systems with this behaviour include water + ethylene glycol and tin - magnesium.

Use the result of the previous problem to calculate the freezing temperature of seawater.

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