How many grams of KH2PO4 are required to make a 20.0 mmol/L phosphoric acid buffer, pH 3.50?

To determine how many grams of KH2PO4 are needed to prepare a buffer solution with a specific molarity, it's essential to understand the relationship between the buffer components and their concentrations. In this scenario, you're tasked with making a 20.0 mmol/L phosphoric acid buffer at a pH of 3.50, and KH2PO4 is a part of this buffer system.

The pH of the buffer indicates that it is being created in an acidic environment using phosphoric acid and its conjugate base. In this case, KH2PO4 acts as the acid, and K2HPO4 would serve as the conjugate base. The required concentration of 20.0 mmol/L suggests that you need to have sufficient amounts of both components.

Now, the first step to calculate the mass of KH2PO4 needed involves converting the desired molarity into moles. Since you want 20.0 mmol/L (millimoles per liter), this amounts to 0.0200 moles per liter.

Next, to find the mass, you need to utilize the molar mass of KH2PO4, which is approximately 136.09 g/mol. Thus, you can calculate the grams needed for the buffer