Please explain as you go so I can fully understand. 1) If a 10.0 L vessel is cha
ID: 907412 • Letter: P
Question
Please explain as you go so I can fully understand.
1) If a 10.0 L vessel is charged with 0.482 mol N2, 0.933 mol O2, calculate the concentration of each species at equilibrium.
2 N2 + O2 2 N2O Kc = 2.0 x 10-32
2) Calculate the equilibrium concentrations of reactants and products of a 1.00 L vessel is charged with 0.0100 M N2 and 0.020 M H2.
3 H2 + N2 2. NH3 Kc = 1.1 x 10-3
4) Calculate the pH of 0.100 M HBr and of 0.100 M KOH.
5) Calculate the pH of 0.100 M acetic acid, Ka = 1.75 x 10-5.
6) Calculate the pH of 0.100 M iodic acid, Ka = 0.17.
Explanation / Answer
2 N2 + O2 2 N2O
initial 0.482 mol 0.933
at eq 0.482-2x 0.933-x 2x
Kc = 2.0 x 10-32 = [N2O ]2/[N2]2[O2]
2.0 x10-32 = 10x [2x]2/(0.482-2x)2 (0.933-x)
solve for x , assuming x to be very small as equilibrium constant is too small,
x = 1.4 x 10-17
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2) 3 H2 + N2 2. NH3
initial .02 .01
at eq .02-3x .01-x 2x
put in equilibrium equation
Kc = 1.1 x 10-3 = [NH3]2/[H2]3[N2] = (2x)2/(.02-3x)3 (0.01-x)
solve for x , we get all concentrations
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3) its a neutral soluytion if equal volume are added so pH = 7
4) its a case of weak acid,
so apply formula pH =1/2(pKa)-1/2(logc)
= 1/2(4.7) -1/2(log.1) = 2.85
5) assuming x% disscoiation,
ka = Cx2/ (1-x ) = .17
on solving x = 46%
where x is the degrre of disscoaition & c is the concentration of acid
[H+] = cx = .1 x.46 = 4.6 x 10-2
pH = -log(4.6 x 10-2 ) calculate