Astudent mixes four reagents together, thinking that the solutions will neutralize each other. the solutions mixed together are 50.0 ml of 0.100 m hydrochloric acid, 100.0 ml of 0.200 m of nitric acid, 500.0 ml of 0.0100 m calcium hydroxide, and 200.0 ml of 0.100 m rubidium hydroxide. is the resulting solution neutral? if not, calculate the concentration of excess h or oh? ions left in solution.

Resulting solution will not be neutral because the moles of ions is greater. The remaining concentration of ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

= 0.200 M

=0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of ion from both the acids

moles of in HCl

if 1 L of solution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 0.1 moles= 0.005 moles    (1L=1000mL)

moles of in HCl = 0.005 moles

Similarliy

moles of in

If 1L of solution= 0.200 moles

Then 0.1L of solution= 0.1 0.200 moles= 0.02 moles

moles of in =0.02 moles

so, Total moles of ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of ion from both the bases

Moles of

1 L of = 0.0100 moles

Then in 0.5 L solution = 0.5 0.0100 moles = 0.005 moles

produces two moles of ions

moles of = 0.005 2= 0.01 moles

Moles of in

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2  0.100 moles = 0.02 moles

Moles of = 0.02 moles

so,Total moles of ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both ions

One mole of ions will combine with one mole of ions, so

Total moles of ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of ions = total moles of ions

0.025 moles  will neutralize the 0.025 moles of

Moles of ions is in excess        (from 1 and 2)

The remaining moles of will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of ions =