Baseline Question: Calculate the pH of a 5.6x10-9M HCl aqueous solution. ANS: pH=6.99 Commentary: Since I know HCl is a strong acid and ionization is 100%, I thought I could find pH as -log(5.6x10-9), however my teacher said that was incorrect . She did a Kw=[OH-][H3O+] to find x. Use image below to see my work. The answer to the question is pH= 6.99. Actual Question:I want to know why we had to do an ICE table and find "x" (another H3O concentration) instead of -log(5.6x10-9).
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
Baseline Question: Calculate the pH of a 5.6x10-9M HCl aqueous solution. ANS: pH=6.99
Commentary: Since I know HCl is a strong acid and ionization is 100%, I thought I could find pH as -log(5.6x10-9), however my teacher said that was incorrect . She did a Kw=[OH-][H3O+] to find x. Use image below to see my work. The answer to the question is pH= 6.99.
Actual Question:I want to know why we had to do an ICE table and find "x" (another H3O concentration) instead of -log(5.6x10-9).
![1. Calculate the pH of a 5.6e-9 M HCl aqueous solution.
I
C
E
H₂O
H₂O
H3O+
5.6e-9
+x
5.6e-9 + x
Kw= [OH-] [H3O+]
1.0x1014(x) (5.6x10° + x)
10시에는
1.0x10-¹4 = 5.6x10-%+x²
0= x² +5.6x15% 16 -1.0x10-14
= PH= -log [H3O+]
= -log
OH
0
+x
(5.6x10-99.82×108)
= 6.99
X
x=9.72×10-8M
[130¹] = [5.6x10²7 x²]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ef2acf2-70fd-40cc-bcbe-08eec48c6b31%2Fc2f7c6b5-80cf-4339-9f1f-5f31c7f1a1fd%2Fu92ak7o_processed.jpeg&w=3840&q=75)
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