1. in a 1x10-6 M solution of Ba(oH)2. Rank the species in order of most concentr
ID: 500817 • Letter: 1
Question
1. in a 1x10-6 M solution of Ba(oH)2. Rank the species in order of most concentrated to least Ba OH. H20. Hao. Ba(OH) 2. How would you classify the difference in pH of 0.1 M of HNO, vs 0.15 M of H CCOOH (with a K, of 1x10-5) A PH ishigher for the solution of HNO3than Hac-cooH because HNO forms an equilibrium with water. pH-3 for HNO3 but pH-1 for H3C-COOH B pH is higher for the solution of Hac-COOH than H H CCOOH forms an equilibrium with water. pH-1 for HNO, but pH for H C-COOH The conjugate base of HNO3 is very strong and changes all the nu mbers around D poH is higher for the solution of HNO than HaCcooH because HN forms an equilibrium with water. pH-3 for HNO but pH-1 for R3C-COOH 3. With a weak acid (pK 45) at pH 33-0. what is the percent ionization (or A) of the weak acid Ka DH,oT-OHANICA1 and that gives 10 si 1o asoo32. only 3% ionized O Ka D,on-ATIDIA1 and that gives 10 s/103 oo32. only 3% isionized as CO KaipHalo CAT/DHA) and that gives 10 s/10 aas oo32. only 97% is ionized O Ka Hao1 DHA]I that gives 10 s/10 s as oo32. only 97% is onized Anand 4. How does the equilibrium shift of H C-CooH H O H,C-coo Hao when H.ccoo Na (product) is added using a kinetics al the rate forward will increase and not be equal to the reverse. there willbeashiftto make more products until the rates become equal. the rate forward will increase and not be equal to the reverse. there willbeashiftto make more until the rates become equal. the rate reverse will increase and not be equal to the forward. there willbeashiftto make more reactants until the rates become equal. the rate reverse will increase and not be equal to the forward. there willbeashiftto make more products until the rates become equal.Explanation / Answer
Q1.
[Ba+2] = 10^-6
[OH-] = 2*[Ba+2] = 2*10^-6
H2O = most abundant, always, since solvent
[H3O+] = Kw/[OH-] = (10^-14)/(2*10^-6) = 5e-9
Ba(OH)2 (s) = assume 9
so:
Clearly, H2O is the most abundant, ignore A an B
Then
OH- since it is the double of Ba+2
then Ba+2
afterwards,
H3O+, since it is less than OH- by definition
finally,
Ba(OH)2 as a solid, since it all dissapeared in solution
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