Final answer:
To find the volume of 0.08892 M HNO3 needed to react completely with 0.2352 g of potassium hydrogen phosphate, calculate the moles of K2HPO4, use the molar ratios from the balanced equation to find the moles of HNO3, and convert the moles to volume using its molarity. The volume is approximately 30.32 mL.
Step-by-step explanation:
To determine the volume of 0.08892 M HNO3 required to react completely with 0.2352 g of potassium hydrogen phosphate, we need to use the molar ratios and the formula of the balanced chemical equation. First, we calculate the number of moles of potassium hydrogen phosphate using its molar mass. Then, using the molar ratios from the balanced equation, we can determine the moles of HNO3 needed. Finally, we can convert the moles of HNO3 to volume using its molarity.
The molar mass of K2HPO4 is 174.18 g/mol. So, the moles of K2HPO4 can be calculated as:
moles of K2HPO4 = mass / molar mass = 0.2352 g / 174.18 g/mol = 0.001349 mol
From the balanced equation, we can see that the ratio between K2HPO4 and HNO3 is 1:2. So, the moles of HNO3 needed are:
moles of HNO3 = 2 * moles of K2HPO4 = 2 * 0.001349 mol = 0.002698 mol
The volume of 0.08892 M HNO3 can be calculated using the equation:
volume (in liters) = moles / molarity = 0.002698 mol / 0.08892 mol/L = 0.03032 L = 30.32 mL
Therefore, the volume of 0.08892 M HNO3 required to react completely with 0.2352 g of potassium hydrogen phosphate is approximately 30.32 mL.