a student mixes 4 reagents together thinking that the solution will neutralize each other . The solution is mixed together are 50 ml of 0.100 M of HCL and 100 ml of 0,200 M of nitric acid 500 ml of 0.0100 M ca hydroxide and 200 ml of 0.100 M rubidium hydroxide . Did the acid bases exactly neutralize each other . If not calculate the concentration of excess H+ or OH-

H+=.05*1+.1*.2-.5*.01*2-.2*.1=.04moles

check that.

The result is neutral

To determine if the acids and bases exactly neutralize each other, we need to calculate the net ionic equation for the reaction and check if all the H+ ions from the acids are consumed by the OH- ions from the bases.

First, let's write the balanced chemical equation for the reaction between the acids (HCl and nitric acid) and the bases (Ca(OH)2 and rubidium hydroxide):

HCl + Ca(OH)2 → CaCl2 + H2O
HNO3 + RbOH → RbNO3 + H2O

Since HCl is a strong acid, it will completely ionize in solution, so we have:

H+ + Cl- + Ca2+ + 2OH- → Ca2+ + 2Cl- + H2O

Since HNO3 is also a strong acid, it will react in a similar way:

H+ + NO3- + Rb+ + OH- → Rb+ + NO3- + H2O

From the balanced equations, we can see that the H+ ions from the acids react with the OH- ions from the bases to form water. This implies that if all the H+ ions have reacted, we should have equal amounts of H+ and OH- ions, resulting in a neutral solution.

Now we can calculate the concentration of excess H+ or OH- ions if they are not completely neutralized. We'll compare the amounts of H+ and OH- ions in the solution.

For the H+ ions:
Initial moles of H+ = initial volume (in L) × initial concentration (in M)
Moles of H+ = 50 ml × 0.1 M + 100 ml × 0.2 M = 5 mmol + 20 mmol = 25 mmol = 0.025 moles

For the OH- ions:
Initial moles of OH- = initial volume (in L) × initial concentration (in M)
Moles of OH- = 500 ml × 0.01 M + 200 ml × 0.1 M = 5 mmol + 20 mmol = 25 mmol = 0.025 moles

Since the moles of H+ and OH- are equal, there is no excess of H+ or OH- ions. Therefore, the acid and bases did exactly neutralize each other, resulting in a neutral solution.

Note: It is important to note that this calculation assumes that all reactions proceed to completion and that there are no other reactions occurring. Additionally, the calculation assumes that the volumes can be directly added together since the total solution volume is given. However, in more complex mixtures, the volumes would need to be converted to moles and then potentially back to concentrations to properly account for the diluted or concentrated solutions.