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Given that solubility product of AgCl = 1.8 X 10^-10
Dissociation of AgCl can be represented as follows,
AgCl(s) ↔ Ag+(ag) + Cl-(aq)
Let, [Ag+] = [Cl-] = S
∴Ksp = [Ag+][Cl-] = S^2
∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3
Now, Molarity of solution =
∴ 1.34 x 10^-5 =
∴ Weight of AgCl present in solution = 1.92 X 10^-3 g
Thus, mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g
Dissociation of AgCl can be represented as follows,
AgCl(s) ↔ Ag+(ag) + Cl-(aq)
Let, [Ag+] = [Cl-] = S
∴Ksp = [Ag+][Cl-] = S^2
∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3
Now, Molarity of solution =
∴ 1.34 x 10^-5 =
∴ Weight of AgCl present in solution = 1.92 X 10^-3 g
Thus, mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g
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 =