Answer:
the answe is oviusly A hope this helps
The calculated pH of the solutions are given below:
- The pH of the solution after 14.0 ml of base is added to it is calculated as 1.45.
- The pH of the solution after 19.8 ml of base is added to it is calculated as 3.0
- The pH of the solution after 20.0 ml of base is added to it is calculated as 7.
<h3>What is pH?</h3>
This is the level of acidity or alkalinity of an aqueous solution. The pH of a substances tells if its an acid, base or neutral.
HBr and NaOH while in water would dissociate and they would become H+ and OH- respectively.
Mol of HBr = Mol * Vol
= 0.200 * 20mL
= 4 mmol
A. Moles of NaOH added
= 0.2 X 14
= 2.8 mmol
Moles of H+ that did not react
= 4 - 2.8
= 1.2 mmol
1.2/1000 = 0.0012 moles
Total volume = 20 + 14
= 34 mL to litres
= 0.034 L
Molarity of H+ = 0.0012 / 0.034L
= 0.035 M
-log[0.035] = 1.45
The pH of the solution is 1.45
B. NaOH added = 19.8 * 0.2 =
3.96 mmoles
The unreacted solution
= 4.0 - 3.96
= 0.04 mmol
0.04/1000 = 0.00004 moles
Total volume = 20 + 19.8
= 39.8mL
Converted to litres = 0.0398L
Molarity = 0.00004 / 0.0398L
= 0.001
-log(0.001) = 3.0
The ph Is therefore 3.0
C. Moles of NaOH added = 0.2*20mL = 4mmol
All the H+ are going to be consumed here. This would result into a neutral solution pH = 7
<u>Complete question:</u>
20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the PH of the solution after the following volumes of base have been added.
A. 14.0 mL
B. 19.8 mL
C. 20.0 mL
Read more on pH here; brainly.com/question/22390063
Answer: Option B
Explanation: Passage 1 contains a false dilemma, while passage 2 contains a bandwagon appeal.
