Question

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 9.00 mL pH = 12.73 (b) 29.80 mL pH = 1 (c) 38.00 mL pH = 1.9

Answer 1

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

Answer 2

The study of the chemical and its bonds is called chemistry.

The correct answer is pH = 12.73, 10.52, and 1.93

What is balanced reaction?

• The balanced reaction is stated as the number of moles and atoms present equally at both sides of the reactant and the product.

The balanced reaction equation is as follow:-

• $KOH + HBr ---> H_2O + KBr$

The amount of KOH is present in the equation.

$n = CV = (0.1000)(30.00) = 3.000m\ mole$

The solution of the first question is as follows:-

• The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

$(0.1000)(9.00) = 0.900m\ mole$

• This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

$(3.000m\ mol) - (0.900m\ mol) = 2.100m\ mol KOH$

• After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

The formula used to solve the question is as follows:-

$C = \frac{n}{V} = \frac{(2.100)}{(39.00)} = 0.0538461 M KOH$

The pOH and pH of the solution can then be calculated as follows:-

$pOH = -log[OH^-] = -log(0.0538461) = 1.2688pH = 14 - pOH = 14 - 1.2688 = 12.73$

The answer to the second question is as follows:-

• The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

All the step is repeated same which is done in question 1.

$(0.1000 )(29.80 mL) = 2.980 m\ mol$

$(3.000m\ mol) - (2.980m\ mol) = 0.0200m\ mol KOH$

$C = \frac{n}{V} = \frac{(0.0200)}{(59.80)}\\ = 0.0003344 M KOH$

• The pOH and pH of the solution is as follows:- $pOH = -log[OH^-] = -log(0.0003344) = 3.476pH = 14 - pOH = 14 - 3.476 = 10.52$

The answer of the third answer is as follows:-

• The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

$(0.1000)(38.00 mL) = 3.800 mmol$

$(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr=\\\frac{n}{V}\\\\\\= \frac{(0.800 mmol)} { (68.00 mL)} = 0.01176 M HBr$

The pH of the solution is as follows:-

Hence, the correct answer is for all the formulas is 12.73, 10.52, and 1.93

For more information about the equilibrium constant, refer to the link:-

https://brainly.com/question/10038290