Question

3.)A certain weak acid, HA, with a Ka value of 5.61×10?6, is titrated with NaOH.

Part A

A solution is made by titrating 7.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH?

Express the pH numerically to two decimal places.

Part B

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 35.0 mL ?

Express the pH numerically to two decimal places.

5.) Part A

20 mL of 0.25 M of NH₃ is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. K _b (NH₃) = 1.8 × 10^?5

	A.)12.50
	B.)7.00
	C.)4.74
	D.)1.12

Answer 1

Concepts and reason

The strength of an acid is measured in terms of its ability to release ions in a solution. This is measured by calculating the acid dissociation constant () of the species in a given medium. The dissociation constant is measured for an equilibrium that exists between the undissociated acid and the dissociated acid. The concentrations of the reactants/products quantify the amount of the reactant/product. The dissociation constant’s numerical value involves exponential components. For the convenience of using a new quantity that effectively condenses this numerical form, symbol is introduced. () takes the form of a simple numeral.

Fundamentals

For an acid

The equilibrium that is achieved on its partial dissolution could be written as follows:

The equilibrium constant for the above reaction forms the acid dissociation constant of the species. Hence, could be written as follows:

- The concentration of ions.

- The concentration of ions.

- The concentration of undissociated acid.

The is calculated as follows:

$pH=-log[H] $

… Henderson-Hasselbach relation

is the counterpart of , for a base.

at 298 K and 1 atm pressure.

is the negative logarithm of the ionic product of water.

(3.A)

The reaction between acid and base can be written as

of acid is added with of . Hence, of salt would be formed, of acid would remain, while the entire base would be consumed.

The of the weak acid is given as follows:

Hence, the would be .

Therefore

(3.B)

The of the acid is calculated as 5.25,

The following reaction informs that and form a base-conjugate acid pair.

Hence, the would be.

Therefore, the base dissociation constant would be.

The salt hydrolyses into its constituent ions. The concentration of the salt could be calculated as follows:

Hence, the concentration of the hydroxyl ions generated as a result that could be calculated as follows:

Therefore, $POH=4.72 $

Hence, the pH of the solution is.

(5.A)

The reaction could be given as follows:

1 mol of aqueous ammonia consumes 1 mol of hydrochloric acid. Hence, the addition of of to of aqueous ammonia would generate an excess of protons.

of acid generates of protons.

of base contains of ammonia.

There shall be an excess of protons.

Hence, the concentration of protons could be calculated as follows;

The wrong options are as follows:

The concentration of protons in the solution is.

Hence, the pH of the solution would be

Therefore, the right option would be

D.) 1.12

Ans: Part 3.A

The pH of the solution is 4.47.