Question

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.

Answer 1

Answer: Resulting solution will not be neutral because the moles of
`OH^-`ions is greater. The remaining concentration of
`[OH^-]`ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

`[HNO_3]` = 0.200 M

`[Ca(OH)_2]` =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of
`H^+` ion from both the acids

moles of
`H^+` in HCl

`HCl\rightarrow {H^+}+Cl^-`

if 1 L of
`HCl`solution =0.100 moles of HCl

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

moles of
`H^+` in HCl = 0.005 moles

Similarliy

moles of
`H^+` in
`HNO_3`

`HNO_3\rightarrow H^++NO_3^-}`

If 1L of
`HNO_3` solution= 0.200 moles

Then 0.1L of
`HNO_3` solution= 0.1
`*` 0.200 moles= 0.02 moles

moles of
`H^+` in
`HNO_3` =0.02 moles

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

Step 2: Calculating the total moles of
`[OH^-]` ion from both the bases

Moles of
`OH^-\text{ in }Ca(OH)_2`

`Ca(OH)_2\rightarrow Ca^2{+}+2OH^-`

1 L of
`Ca(OH)_2`= 0.0100 moles

Then in 0.5 L
`Ca(OH)_2` solution = 0.5
`*`0.0100 moles = 0.005 moles

`Ca(OH)_2` produces two moles of
`OH^-` ions

moles of
`OH^-` = 0.005
`*` 2= 0.01 moles

Moles of
`OH^-` in
`RbOH`

`RbOH\rightarrow Rb^++OH^-`

1 L of RbOH= 0.100 moles

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

Moles of
`OH^-` = 0.02 moles

so,Total moles of
`OH^-` ions = 0.01 + 0.02=0.030 moles ....(2)

Step 3: Comparing the moles of both
`H^+\text{ and }OH^-` ions

One mole of
`H^+` ions will combine with one mole of
`OH^-` ions, so

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

Total moles of
`OH^-` ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of
`H^+` ions = total moles of
`OH^-` ions

0.025 moles
`H^+` will neutralize the 0.025 moles of
`OH^-`

Moles of
`OH^-` ions is in excess (from 1 and 2)

The remaining moles of
`OH^-` will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of
`OH^-` ions =
`\frac{\text{Moles remaining}}{\text{Total volume}}`

`[OH^-]=(0.005)/(0.85)=0.0058M`