The pH of a solution of CO 2 at 38 ∘ C in equilibrium has to be calculated Concept Information: Acid ionization constant K a : Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent. The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization. The ionization of a weak acid HA can be given as follows, HA ( a q ) → H + ( a q ) + A - ( a q ) The equilibrium expression for the above reaction is given below. K a = [ H + ][A - ] [ HA] Where, K a is acid ionization constant, [ H + ] is concentration of hydrogen ion [ A - ] is concentration of acid anion [ HA] is concentration of the acid pH definition: The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration. pH = -log[H 3 O + ] To Calculate: The pH of a solution of CO 2 at 38 ∘ C in equilibrium
The pH of a solution of CO 2 at 38 ∘ C in equilibrium has to be calculated Concept Information: Acid ionization constant K a : Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent. The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization. The ionization of a weak acid HA can be given as follows, HA ( a q ) → H + ( a q ) + A - ( a q ) The equilibrium expression for the above reaction is given below. K a = [ H + ][A - ] [ HA] Where, K a is acid ionization constant, [ H + ] is concentration of hydrogen ion [ A - ] is concentration of acid anion [ HA] is concentration of the acid pH definition: The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration. pH = -log[H 3 O + ] To Calculate: The pH of a solution of CO 2 at 38 ∘ C in equilibrium
Solution Summary: The author explains the ionization of a solution of CO_ 2 in equilibrium and the equilibrium expression for the reaction.
PROBLEM 5+
What is the major product of each of the following reactions?
a. CH3CH2CHCH3 + HBr
d.
+ HBr
A
OH
OH
CH3
CH3
e.
b.
-OH + HCI
+ HCl
A,
OH
CH3
OH
CH3
c. CH3C CHCH3 + HBr
CH3 OH
f.
CHCH3 + HCl ^>
One suggestion for solving the fuel shortage due to decreasing volumes of fossil fuels
are hydrogen / oxygen fuel cells.
a. State the two half-cell reaction equations for such fuel cells. Calculate the cell
potential as well as the electrical work gained by this fuel cell at standard conditions
with E002/H20 = 1.229 V.
b. Compare the fuel cell to the Gibbs free energy of the combustion reaction of
n-octane at standard conditions. Use ASºm, n-Oct., 1 = 361.2 J/mol K.
a. Determine the electrochemical potential of the following cell using
E°Mg2+/Mg = -2.362 V.
Mg | Mg2+ (a=104) || H* (a = 4) | H2 (p
= 0.5 bar) | Pt
b. A galvanic chain consists of Co²+ / Co and Ag+ / Ag half-cells with
EºCo²+/Co = -0.282 V and Eº Ag+/Ag = 0.799 V. Determine which half-cell will be
reduced and which one will be oxidised. Furthermore, calculate the electrochemical
potential as well as the equilibrium constant of the whole cell at
i. [Co²+] = 0.1 M and [Ag+] = 0.5 M
ii. [Co²+] = 0.001 M and [Ag*] = 1.5 M
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell