It’s on September 12th :)
C should be the answer to the following formula
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
<u>Answer:</u> The pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.
<u>Explanation:</u>
There are three types of solution: acidic, basic and neutral
To determine the type of solution, we look at the pH values.
- The pH range of acidic solution is 0 to 6.9
- The pH range of basic solution is 7.1 to 14
- The pH of neutral solution is 7.
We are given:
Concentration of HI = 0.100 M
1 mole of HI produces 1 mole of hydrogen ions and 1 mole of iodide ions
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=0.100M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.100M)
Putting values in above equation, we get:

To calculate the pOH of the solution, we use the equation:
pH + pOH = 14

Hence, the pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.