Calculate the volume of 2.503 N HNO3 added to 100 mL of HCl in order to prepare 1.1251 N acid solution from the following data: 58.96 mL NaOH will neutralize 2.521g of potassium tetroxalate and 45.57 mL of HCl will neutralize 53.0 mL of the NaOH solution.
To calculate the volume of 2.503 N HNO3 needed to prepare a 1.1251 N acid solution, we can follow these steps:
Step 1: Determine the concentration of the NaOH solution.
We know that 58.96 mL of NaOH neutralizes 2.521 g of potassium tetroxalate. From this information, we can calculate the concentration of NaOH as follows:
Concentration of NaOH = (grams of potassium tetroxalate / volume of NaOH solution in liters)
First, we need to convert the volume of NaOH solution to liters:
58.96 mL = 0.05896 L
Now, we can calculate the concentration of NaOH:
Concentration of NaOH = (2.521 g / 0.05896 L)
Step 2: Determine the number of moles of NaOH used to neutralize HCl.
We know that 45.57 mL of HCl neutralizes 53.0 mL of the NaOH solution. From this information, we can calculate the number of moles of NaOH used to neutralize HCl:
Moles of NaOH = Concentration of NaOH × Volume of NaOH solution used (in liters)
Moles of NaOH = Concentration of NaOH × (53.0 mL / 1000)
Step 3: Calculate the volume of HNO3 required to prepare the 1.1251 N acid solution.
Now we can use the stoichiometry of the balanced chemical equation to determine the volume of HNO3 required.
The balanced chemical equation for the neutralization reaction between HNO3 and NaOH is:
HNO3 + NaOH → NaNO3 + H2O
From the balanced equation, we can see that one mole of HNO3 reacts with one mole of NaOH.
So, the volume of HNO3 needed can be calculated as follows:
Volume of HNO3 = Moles of NaOH × (Volume of HCl / Moles of HCl)
However, we haven't been provided with the concentration or volume of HCl. Please provide that information so we can proceed with the calculation.