To calculate the pH of the resulting solution when 85 mL of 0.3 M nitric acid is mixed with 75 mL of 0.2 M magnesium hydroxide, we need to determine the concentration of the excess H+ or OH- ions after the reaction between the two solutions has occurred.
First, we need to write the balanced chemical equation for the reaction:
HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O
This equation tells us that 1 mole of HNO3 reacts with 1 mole of Mg(OH)2 to produce 1 mole of Mg(NO3)2 and 2 moles of H2O.
Next, we need to determine which reactant is limiting and which is in excess. To do this, we can use the balanced chemical equation to calculate the number of moles of each reactant:
Number of moles of HNO3 = (0.3 mol/L) x (0.085 L) = 0.0255 mol
Number of moles of Mg(OH)2 = (0.2 mol/L) x (0.075 L) = 0.015 mol
Since the number of moles of Mg(OH)2 is less than the number of moles of HNO3, Mg(OH)2 is the limiting reactant and HNO3 is in excess.
The Mg(OH)2 will react completely with the HNO3 to form Mg(NO3)2 and H2O. The balanced chemical equation tells us that 1 mole of Mg(OH)2 reacts with 2 moles of HNO3 to produce 1 mole of Mg(NO3)2 and 2 moles of H2O. Therefore, all of the Mg(OH)2 will react with 0.015 moles of the HNO3, leaving 0.0105 moles of HNO3 in excess.
The concentration of the excess HNO3 can be calculated as follows:
Concentration of excess HNO3 = (0.0105 mol) / (0.085 L + 0.075 L) = 0.06 M
Now, we can use the concentration of the excess H+ ions to calculate the pH of the resulting solution. Since HNO3 is a strong acid, it will completely dissociate in water to form H+ and NO3- ions. Therefore, the concentration of H+ ions in