This article discusses the process of neutralizing the pH of water using vinegar. It explains that to reduce the pH of water from 8 to 7, an acid (vinegar) is added.
The volume of vinegar = (8 - 7) x (500 mL) x (10^-14) » ÷ (0.086 M) x (1.8 x 10^-5) » = 3.22 mL
This means that you need to add about 3.22 mL of vinegar to 500 mL of water to neutralise it to a pH of 7.
The volume of vinegar = (8 - 7) x (500 mL) x (10^-14) » ÷ (0.086 M) x (1.8 x 10^-5) » = 3.22 mL
This means that you need to add about 3.22 mL of vinegar to 500 mL of water to neutralise it to a pH of 7.
