Solve the following problems. 1. The concentration of H+ ions in a bottle of table wine was 3.2 x 10-4 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a H+ion concentration equal to 1.0 x 10-3 M. Calculate the pH of the wine on each of these occasions. 2. The pH of rainwater collected in ANHS on a particular day was 4.82. Calculate the H+ ion concentration of the rainwater. 3 Find the nH if the [OH-] is equal to 10x 10-6
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.
(C3)
Hello! I just want to ask for help whether the answers in the given pictures are correct. If it's not, please help me recheck and resolve it. Please refer to the given pictures below for the answers and questions. Please read the instructions and directions very carefully. Double and triple check your answers, previous tutors got it wrong.
NOTE: Type only your answers. Please do not handwritten your answers. Make sure your formulas, solutions and answers' format are all correct.
![1. Given [H*]= 3.2 x 10-4 M, Find PH
Given [H] = 1.0 x 10-³ M,
-3 Find PH
2. Given: PH=4.82, Find [H*]
3. Given: [OH-] = 1.0 x 10-6 M, Find PH
Step 2
1. (a) On first ocassion:
PH= -log₁0[H+] = -log₁0[3.2 x 10-4] = 3.5 (Answer)
(a) On second ocassion:
PH= -log₁0[H*] = -log₁0[1.0 x 10-³] = 3.0 (Answer)
2) PH= -log₁0[H*]
4.82 = -log10[H*]
[H] = 10-4.82 1.51 X 10-5 M (Answer)
=
3)
Given: [OH-] = 1.0 x 10-6 M
POH = -log₁0[OH-] = -log₁0[1.0 X 106] = 6
PH+POH =14
PH + 6 = 14
PH = 14-6-8 (Answer)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92a0eb1b-c8e1-49c9-9609-54c6a81f749c%2F5a55150b-2692-43c9-b642-cf0e5b80ddba%2Ffkno9jr_processed.jpeg&w=3840&q=75)
![Solve the following problems.
1. The concentration of H+ ions in a bottle of table wine was 3.2 x 10-4 M right after the cork was removed. Only half of
the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a H+ion
concentration equal to 1.0 x 10-3 M. Calculate the pH of the wine on each of these occasions.
2. The pH of rainwater collected in ANHS on a particular day was 4.82. Calculate the H+ ion concentration of the
rainwater.
3. Find the pH if the [OH-] is equal to 1.0 x 10-6 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92a0eb1b-c8e1-49c9-9609-54c6a81f749c%2F5a55150b-2692-43c9-b642-cf0e5b80ddba%2Fgow7esa_processed.jpeg&w=3840&q=75)
Step by step
Solved in 5 steps
