1.) Assuming complete dissociation, what is the pH of a 3.51 mg/L Ba(OH)2 soluti
ID: 819777 • Letter: 1
Question
1.) Assuming complete dissociation, what is the pH of a 3.51 mg/L Ba(OH)2 solution?**Is there a specific formula for this, please help!
2.) Enough of a monoprotic acid is dissolved in water to produce a 0.0177 M solution. The pH of the resulting solution is 2.32. Calculate the Ka for the acid.
**For Ka I got .56 and then converting that into pK I got .252, but I don't believe this is right
1.) Assuming complete dissociation, what is the pH of a 3.51 mg/L Ba(OH)2 solution?
**Is there a specific formula for this, please help!
2.) Enough of a monoprotic acid is dissolved in water to produce a 0.0177 M solution. The pH of the resulting solution is 2.32. Calculate the Ka for the acid.
**For Ka I got .56 and then converting that into pK I got .252, but I don't believe this is right
Explanation / Answer
1.)convert mg/L to g/L (multiplying by "1g/1000mg")
then convert g/L to mol/L (by dividing by the molas mass of Ba(OH)2)
So : 3.51 mg/L /1000 = 0.00351 g/L
0.00351 g/L / 171.34 g/mol = 2.05x10^-5 M
Ba(OH)2 -> Ba^2+ + 2OH-
Multiply the molarity by two (since when it dissociate, you get 2 OH- ions per Ba(OH)2)
2.05x10^-5 M x 2 = 4.1x10^-5 M
pH= 14- pOH
pOH= -log [OH-]
pOH = -log[4.1x10^-5 M] = 4.39
pH = 14 - 4.39 = 9.61
Therefore the pH is 9.61
2.)[H+]= 10^-2.32 =0.00478 M
HA <=> H+ + A-
start
0.0177
change
-x .. . . +x . . .+x
at equilibrium
0.0177-x. . . x. . .x
x = 0.00478 M= [H+]= [A-]
[HA]= 0.0177 - 0.00478=0.0129 M
Ka = [H+][A-]/ [HA]= ( 0.00478)(0.00478)/ 0.0129= 1.77 x 10^-3
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