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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.

User Fordareh
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1 Answer

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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
HClsolution =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

User Phimath
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