A) CH3NH2 (methyl amine) is a weak base. Give the correct chemical equation for the aqueous base dissociation reaction.
CH3NH2(aq) + H2O(l) ⇋ CH3NH3+ (aq) + HO−(aq) |
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CH3NH2(aq) + H2O(l) ⇋ CH3NH3+ (aq) + H3O+(aq) |
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CH3NH2(aq) + H3O+(aq) ⇋ CH3NH3+ (aq) + H2O−(l) |
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CH3NH2(aq) + H2O(l) ⇋ CH3NH− (aq) + H3O+(aq) |
B) What concentration of ammonia is required to have a solution with a pH of 11.23? Kb = 1.8x10-5
A) CH3NH2 (methyl amine) is a weak base. Give the correct chemical equation for the aqueous...
A 25.00 mL sample of a 0.250 M aqueous solution CH3NH2 (a weak base) is titrated with an 0.100 M aqueous solution of HCl (a strong acid.) The molecular and net ionic equation for the reaction is provided below. The Kb value used for CH3NH2 is 4.4x10^-4. Find the pH of the solution after addition of 15.00 mL of the aqueous solution of HCl Molecular: CH3NH2 (aq) + HCl (aq) → CH3NH3+ + Cl— Net ionic: CH3NH2 (aq) + H+ →...
The weak base methylamine, CH3NH2, has Kb= 4.2x10^-4. CH3NH2 + H2O -> CH3NH3+ + OH- Calculate the equilibrium hydroxide ion concentration in a 0.95 M solution of the base. [OH-] = What are the pH and pOH of the solution?
Methylamine, CH3NH2, is a weak base and was featured in the TV show Breaking Bad. A 0.504 M solution of methylamine has equilibrium concentrations of both CH3NH,* and OH' equal to 0.0150 M. What is the Kb of the base? CH3NH2 (aq) + H20 (1) = CH3NH3* (aq) + OH (aq)
Methylamine, CH3NH4, is a weak base. A 0.504 M solution of methylamine has equilibrium concentrations of both CH3NH3^+ and OH^- equal to 0.0150 M. What is the Kb of the base? CH3NH2(aq) + H2O(l) <> CH3NH3^+(aq) + OH^-(aq)
In aqueous solution, the azide ion, N3−, is a weak base that accepts a proton from water to form the hydroxide ion, OH −, and hydrazoic acid, HN3, according to the following equation. N3−(aq) + H2O(l) equilibrium reaction arrow OH −(aq) + HN3(aq) The base-dissociation constant (Kb) for this base is 4.04 ✕ 10−10. If a 0.082 M solution of azide ions is prepared, what is the final pH of the solution? (Assume that the temperature is 25°C.)
Determine the ammonia concentration of an aqueous solution that
has a pH of 11.30. The equation for the dissociation of
NH3 (Kb = 1.8 × 10-5) is
below:
Determine the ammonia concentration of an aqueous solution that
has a pH of 11.30. The equation for the dissociation of
NH3 (Kb = 1.8 × 10-5) is
below:
a)9.0 × 10-3 mol L-1
b)2.7 mol L-1
c)0.22 mol L-1
d)2.0 × 10-3 mol L-1
NH3(aq) + H20(1) = NH4+(aq) + OH-(aq) NH3(aq)...
A) Calculate the pH of a 0.0116 M aqueous solution of methylamine (CH3NH2, Kb = 4.2×10-4) and the equilibrium concentrations of the weak base and its conjugate acid. pH = ? [CH3NH2]equilibrium = ? M [CH3NH3+ ]equilibrium = ? M B)Calculate the pH of a 0.0115 M aqueous solution of nitrous acid (HNO2, Ka= 4.6×10-4) and the equilibrium concentrations of the weak acid and its conjugate base. pH = ? [HNO2]equilibrium = ? M [NO2- ]equilibrium = ? M C)...
4. Use of the base methyl amine (CH3NH2) to replace ozone-damaging compounds is becoming more widespread. Methyl amine is now used to make pesticides and herbicides to protect potatoes and tomatoes. The K) for CH3NH2 is 4.2 x 10". (A) What would be the pH of a 0.0010 M solution of this base?! Solution:
Which of the following is the correct chemical equation for the "Kb" reaction for the weak base C10H7O− (2-naphthoxide ion)? A) C10H7O− + H3O+ ⇌ C10H7OH + H2O B) C10H7OH + H2O ⇌ C10H7O− + H3O+ C) C10H7O− + H+ ⇌ C10H7OH D) C10H7OH + OH− ⇌ C10H7O− + H2O E) C10H7O− + H2O ⇌ C10H7OH + OH− Given that the Ka value for 2-naphthol (C10H7OH) is 2.7×10-10, what is the value of Kb for the conjugate base C10H7O−?
The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) this constant is given by Kb=[BH+][OH−][B] Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value. Percent ionization=[OH−] equilibrium[B] initial×100% Strong bases, for which Kb is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization....