1. 78.4 mL of a 0.515 M solution of HNO$_3$ is titrated with Fe(OH)$_3$. It takes 45.2 mL of Fe(OH)$_3$ to reach the equivalence point. What is the concentration (molarity) of Fe(OH)$_3$? Answer to 2 decimal places.

2. 66.9 mL of a 0.447 M solution of H$_3$PO$_4$ is titrated with Ca(OH)$_2$. It takes 38.7 mL of Ca(OH)$_2$ to reach the equivalence point. What is the concentration (molarity) of Ca(OH)$_2$? Answer to 2 decimal places.

3. 77.4 mL of a 0.13 M solution of HCl is titrated with Ca(OH)$_2$. It takes 56.5 mL of Ca(OH)$_2$ to reach the equivalence point. What is the concentration (molarity) of Ca(OH)$_2$? Answer to 2 decimal places.

To find the concentration (molarity) of Fe(OH)3 in a titration, we can use the equation: Moles of Fe(OH)3 = Moles of HNO3. This process can be similarly applied to the other two titration problems to find the concentrations of Ca(OH)2.

Explanation:

To find the concentration (molarity) of Fe(OH)3 in a titration, we can use the equation:

Moles of Fe(OH)3 = Moles of HNO3

Given that the volume and molarity of HNO3 is 78.4 mL and 0.515M respectively, and the volume of Fe(OH)3 is 45.2 mL, we can calculate the moles of HNO3 and then use it to find the concentration of Fe(OH)3.

Moles of HNO3 = volume of HNO3 (L) x molarity of HNO3 (mol/L)

Moles of Fe(OH)3 = Moles of HNO3

Molarity of Fe(OH)3 = Moles of Fe(OH)3 / Volume of Fe(OH)3 (L)

Plugging in the values, we find the concentration of Fe(OH)3 to be 0.798 M.

This process can be similarly applied to the other two titration problems to find the concentrations of Ca(OH)2. The concentrations are 0.672 M and 0.207 M respectively.

Learn more about Titration here:

https://brainly.com/question/31271061

#SPJ11