2 replies to this topic

[Chemed]

I have a 55 000L body of water, and 9 mol/L concentration of hydrochloric acid. How much acid do I need to add to change the pH from 8.1 to 7.2? Please show calculations.

[breizh]

Chemed ,
knowing that ph =-log10(H+) should be a good start to solve your query .

Acidity of the liquid :
Initial : H+ = 10^-8.1 mole/l before acidification
Final : H+ = 10^-7.2 mole/l after acidification

Mass balance

Initial : 55 000 liters
Final : 55 000 liters +Y l acid

H+ Balance

(550000 + y) * 10^-7.2 - 55000 * 10^-8.1 = y *9 >>>>>then you can calculate y

Hope this helps
Breizh

[Adam_K]

Hi Chemed,

Breizh is completely right for pure water, however impure water may contain buffers which will change the response of the water to addition of acids and bases. Is this a theoretical question or are you actually attempting to change the pH of the body of water?

In order to test the response of the water to acid addition, it is necessary to construct a pH curve.

• Start off by collecting a water sample (eg. 100 mL) and transferring this to a beaker (eg. 250 mL).
• Immerse a pH probe in the water to determine the initial pH.
• From a burette (10 - 50 mL capacity), add just enough acid to change the pH by 0.1. Repeat until you reach pH 7 or below.
• Plot the a curve of the amount of acid added and the resulting pH
• Calculate the amount of acid required to change the pH to 7.2. Multiply this value by 10 to find the amount of acid required per litre of water; from this you can then multiply by the volume of the body of water (55,000 L in this case) to find the total acid required.

I would suggest diluting the acid 100 x, ie. take 10 mL, place it in a 1 L volumetric flask and fill to the mark with distilled water. This is because you will require very little of the original acid to neutralise the water, and will therefore get results with a low accuracy.

You can also perform the titration in one go (simply adding enough acid to reach pH 7.2) however this won't allow you to understand the response of the water to the acid.

Regards,
Adam

Chemed ,
knowing that ph =-log10(H+) should be a good start to solve your query .

Acidity of the liquid :
Initial : H+ = 10^-8.1 mole/l before acidification
Final : H+ = 10^-7.2 mole/l after acidification

Mass balance

Initial : 55 000 liters
Final : 55 000 liters +Y l acid

H+ Balance

(550000 + y) * 10^-7.2 - 55000 * 10^-8.1 = y *9 >>>>>then you can calculate y

Hope this helps
Breizh