Acid Base Titrations

We are given three carboxylic acids and their corresponding pKa values, and we need to correctly match them.

Step 1: Understanding pKa values
The pKa value of an acid indicates its acidity; lower pKa values correspond to stronger acids.
- Acetic acid (CH COOH) has a known pKa value of 4.76.
- Benzoic acid (C H COOH) has a pKa value of 4.19.
- Trifluoroacetic acid (CF COOH) is a highly acidic carboxylic acid due to the strong electron-withdrawing effect of fluorine atoms, and its pKa value is 0.23.
Step 2: Matching the values
From the given pKa values, we match:
- (CH COOH) (4.76)
- (C H COOH) (4.19)
- (CF COOH) (0.23)
Thus, the correct matching is A-II, B-III, C-I, which corresponds to option (2).

Step 1: Understanding the Given Data
We are given the standard enthalpies of formation ( ): We need to determine the enthalpy of formation of , denoted as .

Step 2: Applying Hess's Law
From the reaction: Using the enthalpy change equation:

Step 3: Solve for

Step 4: Verify the Correct Answer
Thus, the value of is kJ mol , which matches Option (3).

Step 1: Calculate the Moles of HCl and NaOH

Step 2: Determine the Excess Acid or Base
Since HCl and NaOH react in a 1:1 ratio:

Step 3: Calculate the Final Concentration of
Total volume after mixing:

Step 4: Calculate the pH

Step 5: Verify the Correct Answer
Thus, the final pH of the solution is , which matches Option (2).

Step 1: Understanding Acidic Cleavage of Ethers
- Ethers react with hydrogen bromide (HBr) via acidic cleavage, forming alkyl and aryl bromides.
- The cleavage mechanism depends on the stability of carbocations formed during the reaction.
Step 2: Cleavage of Ether A (Phenyl Ethyl Ether)
1. The ether oxygen is protonated by HBr, making it a better leaving group.
2. Since phenyl carbocation is unstable, the bond breaks to form:
- Phenyl bromide ( ) (as benzyl carbocation is not formed).
- Ethanol ( ), which further reacts with HBr to form ethyl bromide ( ).
Step 3: Cleavage of Ether B (tert-Butyl Propyl Ether)
1. The tert-butyl carbocation ( ) is highly stable, so the ether undergoes SN1 cleavage.
2. This results in:
- tert-Butyl bromide ( ) (from stable carbocation).
- Propyl bromide ( ) (via nucleophilic substitution).
Step 4: Evaluating the Given Options
- Option (1): Incorrect, as the structures do not match the cleavage products.
- Option (2): Incorrect, as it does not correctly represent the products.
- Option (3): Incorrect, as tert-butyl bromide formation is missing.
- Option (4): Correct, as it correctly represents phenyl bromide and ethyl bromide for A, and tert-butyl bromide and propyl bromide for B.
Thus, the correct answer is Option (4).

Step 1: Understanding Beryllium Alloys
Beryllium forms several alloys, but one of the most notable is beryllium copper ( ). This alloy is widely used due to its high strength, good electrical conductivity, and non-magnetic properties.

Step 2: Properties and Applications of Beryllium Copper
- Beryllium copper ( ) is a non-sparking, non-magnetic alloy with excellent strength and corrosion resistance. - It is used in the manufacture of high-strength springs, precision instruments, and aerospace components. - The combination of beryllium with copper enhances mechanical strength while maintaining electrical conductivity.

Step 3: Evaluating the Given Options
- Option (1) Al: Beryllium-aluminum alloys exist but are not commonly used for high-strength springs.
- Option (2) Zn: Zinc alloys do not typically involve beryllium in significant amounts for mechanical applications.
- Option (3) Cu: Correct, as beryllium copper is used in high-strength springs.
- Option (4) Cr: Chromium forms strong alloys but is not the primary metal used with beryllium for this application.
Thus, the correct answer is

Option (3).