The K a of a 0 .400 M formic acid HCOOH solution has to be calculated Concept Information: Acid ionization constant K a : The equilibrium expression for the reaction HA ( a q ) → H + ( a q ) + A - ( a q ) 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 Base ionization constant K b The equilibrium expression for the ionization of weak base B will be, B ( a q ) + H 2 O ( l ) → HB + ( a q ) + OH - ( a q ) K b = [ HB + ] [ OH - ] [ B ] Where K b is base ionization constant, [ OH − ] is concentration of hydroxide ion, [ HB + ] is concentration of conjugate acid, [ B] is concentration of the base Relationship between K a and K b K a × K b =K w
The K a of a 0 .400 M formic acid HCOOH solution has to be calculated Concept Information: Acid ionization constant K a : The equilibrium expression for the reaction HA ( a q ) → H + ( a q ) + A - ( a q ) 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 Base ionization constant K b The equilibrium expression for the ionization of weak base B will be, B ( a q ) + H 2 O ( l ) → HB + ( a q ) + OH - ( a q ) K b = [ HB + ] [ OH - ] [ B ] Where K b is base ionization constant, [ OH − ] is concentration of hydroxide ion, [ HB + ] is concentration of conjugate acid, [ B] is concentration of the base Relationship between K a and K b K a × K b =K w
Solution Summary: The author explains that the equilibrium expression for the ionization of weak base B is given below.
Definition Definition State where the components involved in a reversible reaction, namely reactants and product, do not change concentration any further with time. Chemical equilibrium results when the rate of the forward reaction becomes equal to the rate of the reverse reaction.
Chapter 15, Problem 15.138QP
Interpretation Introduction
Interpretation:
The Ka of a 0.400 M formic acid HCOOH solution has to be calculated
Concept Information:
Acid ionization constantKa:
The equilibrium expression for the reaction HA(aq)→H+(aq)+A-(aq) is given below.
Ka=[H+][A-][HA]
Where Ka is acid ionization constant, [H+] is concentration of hydrogen ion, [A-] is concentration of acid anion, [HA] is concentration of the acid
Base ionization constantKb
The equilibrium expression for the ionization of weak base B will be,
B(aq)+H2O(l)→HB+(aq)+OH-(aq)
Kb=[HB+][OH-][B]
Where Kb is base ionization constant, [OH−] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base
Please help with number 6 I got a negative number could that be right?
1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.
Give the product of the bimolecular elimination from each of the isomeric halogenated compounds.
Reaction A
Reaction B.
КОВ
CH₂
HotBu
+B+
ко
HOIBU
+Br+
Templates More
QQQ
Select Cv Templates More
Cras
QQQ
One of these compounds undergoes elimination 50x faster than the other. Which one and why?
Reaction A because the conformation needed for elimination places the phenyl groups and to each other
Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other.
◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other.
Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.