0.233 mol/0.1 L = 0.0233 M But I get 2.33 !! that will change the pH. Please give me my question back if the previous answer was wrong. Step 1 Given that, Percentage of acetic acid solution = 14% It means, 14 g of acetic acid dissolved in 100 g or 100 mL of water. Moles of acetic acid in 14 g = 14 g/60.052 g/mol = 0.233 mol Molarity of 0.233 mol of 100 mL acetic acid solution = 0.233 mol/0.1 L = 0.0233 M Ka value of acetic acid = 1.7×10-5 Step 2 Now, acetic acid dissociates in water as follows; CH3COOH <----> CH3COO- + H+ Ka = [CH3COO- ][H+]/[CH3COOH] Let, [CH3COO- ] = [H+] = x Then we have, 1.7×10-5 = x.x/0.0233M x = 0.629×10-3 M Now, pH of the acetic acid solution will be pH = -log[H+] = -log0.629×10-3 M = 2.8
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.
0.233 mol/0.1 L = 0.0233 M
But I get 2.33 !! that will change the pH.
Please give me my question back if the previous answer was wrong.
Given that,
Percentage of acetic acid solution = 14%
It means, 14 g of acetic acid dissolved in 100 g or 100 mL of water.
Moles of acetic acid in 14 g = 14 g/60.052 g/mol
= 0.233 mol
Molarity of 0.233 mol of 100 mL acetic acid solution = 0.233 mol/0.1 L = 0.0233 M
Ka value of acetic acid = 1.7×10-5
Now, acetic acid dissociates in water as follows;
CH3COOH <----> CH3COO- + H+
Ka = [CH3COO- ][H+]/[CH3COOH]
Let, [CH3COO- ] = [H+] = x
Then we have,
1.7×10-5 = x.x/0.0233M
x = 0.629×10-3 M
Now, pH of the acetic acid solution will be
pH = -log[H+]
= -log0.629×10-3 M
= 2.8
Step by step
Solved in 2 steps with 2 images
