2,6 -dimetil nonano 4 etil-2,2dimetil-4-propo:0ctano

A chemist titrates of a hypochlorous acid solution with solution at . Calculate the pH at equivalence. The of hypochlorous acid

const2013 [10]

The question is incomplete, here is the complete question:

A chemist titrates 110.0 mL of a 0.2412 M hypochlorous acid (HCIO) solution with 0.0613 M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of hypochlorous acid is 7.50. Round your answer to 2 decimal places

Answer: The pH of the solution is 10.09

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HClO$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.2412M\\V_1=110.0mL\\n_2=1\\M_2=0.0613M\\V_2=?mL$

Putting values in above equation, we get:

$1\times 0.2412\times 110.0=1\times 0.0613\times V_2\\\\V_2=\frac{1\times 0.2412\times 110.0}{1\times 0.0613}=432.8mL$

At equivalence, the number of moles of acid is equal to the number of moles of base. Also, the moles of salt which is NaClO will also be the same.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For HClO:

Molarity of HClO solution = 0.2412 M

Volume of solution = 110.0 mL

Putting values in equation 1, we get:

$0.2412M=\frac{\text{Moles of HClO}\times 1000}{110}\\\\\text{Moles of HClO}=\frac{(0.2412\times 110)}{1000}=0.026532mol$

For NaClO:

Moles of NaClO = 0.026532 moles

Volume of solution = [432.8 + 110] mL = 542.8 mL

Putting values in above equation, we get:

$\text{Molarity of NaClO}=\frac{0.026532\times 1000}{542.8}=0.0489M$

To calculate the pH of the solution, we use the equation:

$pH=7+\frac{1}{2}[pK_a+\log C]$

where,

$pK_a$ = negative logarithm of weak acid which is hypochlorous acid = 7.50

C = concentration of the salt = 0.0489 M

Putting values in above equation, we get:

$pH=7+\frac{1}{2}[7.50+\log (0.0489)]\\\\pH=7+3.09=10.09$

Hence, the pH of the solution is 10.09

4 0

3 years ago

Identify the functional groups present in each molecule.

Colt1911 [192]

Answer:

Carbon 3 is double bonded to an oxygen and attached to carbon 2 and carbon 4. :

Answer: Carbonyl group ( Ketone or aldehyde)

Carbon 17 is attached to an oxygen, which is attached to a hydrogen. :

Answer: Carboxyl group (Carboxylic acid)

A central carbon is attached to an amine, two hydrogens, and a carbon that is double bonded to an oxygen and single bonded to an oxygen attached to a hydrogen. :

Answer: Amide group

An amide group contains both amine and carboxyl

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5 0

2 years ago

What is the Brønsted-Lowry base in this reaction: NH2−+CH3OH→NH3+CH3O−?(1 point)CH3O−cap c cap h sub 3 cap o raised to the negat

Fynjy0 [20]

In the reaction as follows: NH2- + CH3OH → NH3 + CH3O−, NH2- is the Brønsted-Lowry base.

BRØNSTED-LOWRY BASE:

According to Bronsted-Lowry definition of a base and acid, a base is substance that accepts an hydrogen ion or proton (H+) while an acid is a substance that donates a proton.

According to this reaction given as follows: NH2 + CH3OH → NH3+ CH3O-

NH2- is a reactant that accepts a hydrogen ion (H+) to become NH3+
NH3+CH3OH is a reactant that donates hydrogen ion (H+)

Since NH2- accepts a proton, this means that in the reaction as follows: NH2 + CH3OH → NH3 + CH3O−, NH2- is the Brønsted-Lowry base.

Learn more at: brainly.com/question/21736327?referrer=searchResults

3 0

3 years ago

(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM

torisob [31]

Answer:

The answer is " $20 \ mL$ "