The pH is the measurement of the amount of hydrogen ions in a solution. Basic solutions have low concentrations of hydrogen ions, while acids have high concentrations of hydrogen ions. The pH of solutions can be altered by adding acids and bases. Acids will lower the pH while bases will raise the pH. If you blindly mix an acid into water, it is unlikely that you will put in the correct amount. If you put too much acid into a solution, you will have to use a base to raise the pH once again. To avoid wasting acids and bases, you can use a simple calculation to determine exactly how much acid you need to lower water to target the pH level.
Obtain a strong acid. Some examples of strong acids include hydrochloric acid, hydrobromic acid and nitric acid, designated HCl, HBr and HNO_3, respectively. Strong acids are solutions that have an extremely high concentration of hydrogen ions. Hydrogen ions make a solution acidic, while hydroxide ions make a solution basic.
Obtain the concentration of the hydrogen ions, also known as molarity, in your strong acid. If you don’t have the concentration, then you likely have the pH of the solution. If you have the pH, you can convert from pH to molarity by using the following equation:
Molarity = 10^-[pH]
If you have a number more than 1, you likely did this calculation wrong. However, if you have a very strong acid, its pH may be less than zero and yield a concentration more than 1. This resulting value is the molarity of the solution. Molarity is the amount of moles of acid per liter of solution. For example, if your solution has 0.5 molarity, then there is only 0.5 mol of acid per 1 L. Molarity can be calculated using this formula:
Molarity = moles of acid/liter(s) of solution
Obtain the molarity of your water sample using this same method.
Convert your target pH value into molarity using the equation in the previous step.
Calculate how much acid you need to obtain the pH level of your target value. You can calculate this by using the following formula:
M_1V_1 + M_2V_2 = M_3(V_1 + V_2)
In this equation, “M_1” is the molarity of the acid, “V_1” is the volume of the acidic solution, “M_2” is the molarity of the water, and “V_2” is the volume of the water. Converting this equation to solve for “V_1” yields the following equation:
V_1 = (M_3V_2 – M_2V_2)/(M_1 – M_3).