B. Calculate the pH after 17.50 mL of NaOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz • First assume the acid-base neutralization reaction is ireversible and goes to completion: HCH3O2 (ag) + OH" (ag) → H2O () + C2H3O2 (ag) Moles before reaction Change in moles Moles remaining after reaction
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.
Original concentration of Acetic Acid was 0.1033 M and NaOH was 0.1104 M
there is part a done, need to do part B
![A. Calculate the pH of 35.00 mL of acetic acid using the concentration assigned in Beyond Labz:
HC2H3O2 (aa)
2 H* (aa)
+ C2H3O2 (aa)
I
Initial Concentration (M) 0.1033
Change in Concentration (M)
+x
+x
-X-
Equilibrium Concentration (M) 0.1033 - x
E
Equilibrium Concentration,
Calculated (M)
0.1020
1.36x10-3
1.36x10-3
1. Initial concentration [HC2H3O2]0=0.1033 M
2. Ka symbolic expression: [H*][ C2H3O2 ]
[HC2H3O2]
(x)(x)
3. Ka numerical equation with x: 1.8 x 10-S =
0.1033-x
4. Quadratic Formula Method
a. Quadratic equation: x2+ 1.8 x 10-S – 1.86 x 10-6 = 0
-1.8 x 10A-5t/(-1.8x 10)-4(1)(1.86 x 10A-6)
b.Quadratic formula: x =
2(1)
5. Assumption (5% rule) Method
(1.8x10)(0.1033)
a. Equation for x: x =
= 1.36x10-°M
b.% ionization = (1.36x10^-3M)(0.1033M)x100 = 1.3%
6. Equilibrium concentrations
a. [HC2H3O2lm =0.1033 - 1.36x103 = 0.1020M
b.[H* Jeg =1.36x10-3 M
c.[C2H3O2] =1.36x103 M
7. pH = - log(1.36x10-3 ) = 2.8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f55c28e-71db-47ea-b0ac-f3ad1a1e82a5%2Fcd3950f5-64df-4573-a266-3d81b5ea9570%2Fb9nsyqrw_processed.png&w=3840&q=75)
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