Answer: The final volume of this solution is 0.204 L.
Explanation:
Given: Molarity of solution = 2.2 M
Moles of solute = 0.45 mol
Molarity is the number of moles of solute present divided by volume in liters.
Substitute the values into above formula as follows.
Thus, we can conclude that the final volume of this solution is 0.204 L.
Answer: 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 =