I have briefly described the concept of Molarity, pH and (parts per million) ppm and some ways through which they can be calculated in different conditions.

1. CALCULATION OF [H+]
1. [H+] is the molar concentration of Hydrogen ion in a
solution (per liter).
2. Formula to calculate [H+] ion is
pH= -log[H+]
3. Example : Calculate [H+] when pH is 6
Solution :We know that pH = -log[H+]
[H+] = 10-pH
i.e., [H+] = 10-6 is the answer

2. CONCEPT OF MOLARITY
1. Molarity in other term is similar to
concentration and is defined as the moles
per unit volume.
2. There are two different formulas to
calculate molarity.
3. M=(weight of solute/molecular weight)/
volume of solution in Liters
4. Initial conc. X initialVolume=final conc. X
final volume i.e. M1V1 = M2V2

3. Q. What amount(volume) of acid is
required to decrease the pH of a juice
sample (100 Liters) from 5.2 to 3.2 ?
 We know that pH= -log[H+]
 [H+] = 10-pH i.e. [H+] = 10-5.2
[H+] = 6x10-6 i.e., this is the conc. (M1) when the pH
of the juice is 5.2
 [H+] = 10-3.2
[H+] = 4x10-4 i.e., this is the conc. (M2) when the pH
of the juice is 3.2

4. We can calculate the required amount of acid to be
added to the solution
M1 = 6x10-6 , V1 = 100
M2 = 4x10-4 , V2 = ?
Now using the formula M1V1=M2V2
6x10-6 x 100 = 4x10-4 x v2
V2 = 1.5 Liter

5. CONCEPT OF PPM
 ppm stands for parts per million
i.e., 1 ppm= 1 ml in 10-6
 ppm depends on concentration, temperature
etc.

6. Q. What amount of Sodium
Hypochlorite(NaOCl) is required to
prepare 100 liters of a 50 ppm solution
from a 12.5 percent sodium hypochlorite?
 12.5 % of NaOCl solution means
12.5 in 100 units i.e., 12.5 in 102 units
1 unit = 12.5/102
106 unit = (12.5/102)x106 i.e., 125000 ppm
 Final chlorine solution volume = 100 Litres = 100,000 milliliters (ml)
 We an calculate the required amount of NaOCl to be added
M1= 125000 ppm, V1= ?
M2= 50 ppm , V2= 100,000 ml

7. Now using the formula M1V1=M2V2
125000 xV1 = 50 x 100,000
V1 = 40 ml