When salts such as NaCl are added to water, they dissolve and dissociate into io

Question

When salts such as NaCl are added to water, they dissolve and dissociate into ions. NaCl dissolves to Na^+ and Cl^-. Similarly, CH_3NH_3NO_3 is a salt so when it is added to water, it will dissolve and dissociate into ions: CH_3NH_3^+ and NO_3^-. When these ions are in water, what will happen? Will either ion react with water? NO_3^- is the conjugate base of a strong acid, HNO_3 (one of the strong acids from question 3 you have memorized!). As a conjugate base of a strong acid, NO_3^- is very weak and stable so NO_3^- will not react with water. CH_3NH_3^+ is the acid of a weak base, CH_3NH_2 so it will react with water: What is the pH of a 1.2 M CH_3NH_3NO_3 solution? K_b (CH_3NH_2) = 5.2 times 10^-4? 8.68 12.40 1.60 13.92 5.32 Strong bases dissociate completely to give OH^-. Strong bases that you should know are the hydroxides of the first two columns of the periodic table. For example, NaOH, Ca(OH)_2, KOH, etc. Weaker bases are bases where only a FEW molecules of the base dissociate to give OH^-. An example is given below. K_b = [NH_4^+][OH^-]/[NH_3] The pOH of a 0.1 M solution of NaOH is: (Remember that pOH = -log[OH^-]: NH_3 + H_2O rightleftharpoons NH_4^- + OH^- 13.5 0.5 7 1 13

Explanation / Answer

14) I will assume that the volume of the solution is 1L.

CH3NH3NO3 is a salt so it will dissociate completely, so, the amount of moles of CH3NH3NO3 is equal to the amount of moles of CH3NH3+ produced. This means that the concentration of CH3NH3+ is 1.2M.

The reaction of CH3NH3+ with water is the reaction of an acid with water (who acts as a base). This means that we need to find Ka.

Ka x Kb= Kw ----> Ka= 1x10-14/5.2x10-4 = 1.92x10-11

Ka= [H3O+][CH3NH2]/[CH3NH3+]

CH3NH3+ + H2O <--------> CH3NH2 + H3O+

1.2M____________________0_______0

1.2-x____________________x_______x

Ka= x2/1.2-x = 1.92x10-11

x= 4.8x10-6M = [H3O+] ----> pH= -log[H3O+] = 5.32

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16) NaOH is a strong base so it will dissociate completely, [NaOH]=[OH-]= 0.1M

pOH= -log[OH-]= 1 ----> pH + pOH= 14 ----> pH= 13

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