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
It produces a hydrazone in which the carbon is attached with the nitrogen atom with the means of a double bond.
The correct answer is C
We are solving for C so we need to isolate C
Multiply 2 to both sides
2W = CV^2
Divide V^2
C = 2W/V^2
Answer: Objects that are warmer emit more energetic radiation than those that are cooler and so they cool faster. Therefore, in the absence of an external heat ...
Explanation:
Answer: The independent variable is the cause. so choose the cause
Explanation:
The independent variable is the cause.
