Respuesta :
Answer: Resulting solution will not be neutral because the moles of [tex]OH^-[/tex]ions is greater. The remaining concentration of [tex][OH^-][/tex]ions =0.0058 M.
Explanation:
Given,
[HCl]=0.100 M
[tex][HNO_3][/tex] = 0.200 M
[tex][Ca(OH)_2][/tex] =0.0100 M
[RbOH] =0.100 M
Few steps are involved:
Step 1: Calculating the total moles of [tex]H^+[/tex] ion from both the acids
moles of [tex]H^+[/tex] in HCl
[tex]HCl\rightarrow {H^+}+Cl^-[/tex]
if 1 L of [tex]HCl[/tex]solution =0.100 moles of HCl
then 0.05L of HCl solution= 0.05 [tex]\times [/tex]0.1 moles= 0.005 moles (1L=1000mL)
moles of [tex]H^+[/tex] in HCl = 0.005 moles
Similarliy
moles of [tex]H^+[/tex] in [tex] HNO_3[/tex]
[tex]HNO_3\rightarrow H^++NO_3^-}[/tex]
If 1L of [tex]HNO_3[/tex] solution= 0.200 moles
Then 0.1L of [tex]HNO_3[/tex] solution= 0.1 [tex]\times [/tex] 0.200 moles= 0.02 moles
moles of [tex]H^+[/tex] in [tex]HNO_3[/tex] =0.02 moles
so, Total moles of [tex]H^+[/tex] ions = 0.005+0.02= 0.025 moles .....(1)
Step 2: Calculating the total moles of [tex][OH^-][/tex] ion from both the bases
Moles of [tex]OH^-\text{ in }Ca(OH)_2[/tex]
[tex]Ca(OH)_2\rightarrow Ca^2{+}+2OH^-[/tex]
1 L of [tex]Ca(OH)_2[/tex]= 0.0100 moles
Then in 0.5 L [tex]Ca(OH)_2[/tex] solution = 0.5 [tex]\times [/tex]0.0100 moles = 0.005 moles
[tex]Ca(OH)_2[/tex] produces two moles of [tex]OH^-[/tex] ions
moles of [tex]OH^-[/tex] = 0.005 [tex]\times [/tex] 2= 0.01 moles
Moles of [tex]OH^-[/tex] in [tex]RbOH[/tex]
[tex]RbOH\rightarrow Rb^++OH^-[/tex]
1 L of RbOH= 0.100 moles
then 0.2 [RbOH] solution= 0.2 [tex]\times [/tex] 0.100 moles = 0.02 moles
Moles of [tex]OH^-[/tex] = 0.02 moles
so,Total moles of [tex]OH^-[/tex] ions = 0.01 + 0.02=0.030 moles ....(2)
Step 3: Comparing the moles of both [tex]H^+\text{ and }OH^-[/tex] ions
One mole of [tex]H^+[/tex] ions will combine with one mole of [tex]OH^-[/tex] ions, so
Total moles of [tex]H^+[/tex] ions = 0.005+0.02= 0.025 moles....(1)
Total moles of [tex]OH^-[/tex] ions = 0.01 + 0.02=0.030 moles.....(2)
For a solution to be neutral, we have
Total moles of [tex]H^+[/tex] ions = total moles of [tex]OH^-[/tex] ions
0.025 moles [tex]H^+[/tex] will neutralize the 0.025 moles of [tex]OH^-[/tex]
Moles of [tex]OH^-[/tex] ions is in excess (from 1 and 2)
The remaining moles of [tex]OH^-[/tex] will be = 0.030 - 0.025 = 0.005 moles
So,The resulting solution will not be neutral.
Remaining Concentration of [tex]OH^-[/tex] ions = [tex]\frac{\text{Moles remaining}}{\text{Total volume}}[/tex]
[tex][OH^-]=\frac{0.005}{0.85}=0.0058M[/tex]
