Question

180Q1

Part A

Calculate the pH and the concentrations of all species present in 0.14 M H2SO3. (Ka1 = 1.5×10−2, Ka2 = 6.3×10−8)

Express your answer to three significant figures and include the appropriate units.

Part B

Calculate the concentration of H2SO3 in solution.

Express your answer to two significant figures and include the appropriate units.

Part C

Calculate the concentration of HSO−3 in solution.

Express your answer to two significant figures and include the appropriate units.

Part D

Calculate the concentration of H3O+ in solution.

Express your answer to two significant figures and include the appropriate units.

Part E

Calculate the concentration of SO2−3 in solution.

Express your answer to two significant figures and include the appropriate units.

Answer 1

H50,(aq) + H30* (aq) A) Concentration of H SQ = 0.14 M ICE table constructed is as follows; H,80, (aq) + H20(1) F I(M): 0.14 C(M): * E(M) (0.14 - x) 0 + K [ H50,-]H,0+] 1st aci dionization constant, K = ? (H.S03] 1.5x102 - (x)(x) (0.14-x) x= 3.9x102 M From the equilibrium table, (H20"]= x = 3.9x10^M (HSO3 ] = x=3.9x102M [H,803]= (0.14- x) M = {0.14 - 3.9x102) M 0.14M ICE table constructed for the 2nd ionization of H SO, is as follows; HSO, (aq) + H20(1) so,- (aq) + H,0* (aq) I(M): 3.9x10-2 3.9x10-2 C(M) -x +x E(M): (3.9x10-2 - x) 3.9x10-2+x + 2nd acidionization constant, K = [H30*150,3] [HsQ'] 6.3x10 - (3.9x10° +x)(x) (3.9x102 - x) Since H,So, is a weak acid, it ionizes to a little extent So, we can assume that 3.9x102-X 3.9x10-2 3.9x102+x= 3.9x102 & (3.9x10-2)(x) (3.9x102) *=6.3x10* [H30*]= (3.9x10-* +x) M = (3.9x10-2 +6.3x108)M $ 3.9x10^M pH =-log[H,0*] =-log(3.9x10-) = 1.41 B) [H,SO3)= (0.14-x) M =(0.14 -3.9x102) M 0.14M C)From the equilibrium table, (HS0,]= (3.9x10^2 - x)M = (3.9x104 -6.3x108) M * 3.9x10-2M D)[H3O+] = (3.9x10^2 +x) M = (3.9x10-2 +6.3x108)M 3.9x10PM E)[$0,2- ] = (x) M = 6.3x10* M