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A student 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.

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]