A solution of 118 of 0.180 KOH is mixed with a solution of 300 of 0.150 NiSO4. What is the concentration of Ni (2+) that remains in the solution? What is the concentration of SO4 (2+) that...

A solution of 118 of 0.180 KOH is mixed with a solution of 300 of 0.150 NiSO4.

What is the concentration of Ni (2+) that remains in the solution?

What is the concentration of SO4 (2+) that remains in the solution?

What is the  concentration of K (+) that remains in the solution?

How do you find these answers with the given information?

Asked on by mikkiex3

1 Answer | Add Yours

jerichorayel's profile pic

jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on

To solve for the answer, first we have to make the balanced chemical equation. 

`2 KOH (aq) + NiSO_4 (aq) -> Ni(OH)_2 (s) + K_2SO_4 (aq)`

Next, we have to get the number of moles for each reactants.

For KOH:

0.180mol/L x (118/1000)L = 0.02124 moles KOH

For NiSO4:

0.150mol/L x (300/1000)L = 0.0450 moles NiSO4

Then we get to solve for the limiting reagent.

`0.02124 mol es KOH * (1 mol e Ni(OH)_2)/(2 mol es KOH)`

= 0.01062 moles

`0.0450mol es NiSO_4 * (1 mol e Ni(OH)_2)/(1 mol e NiSO_4)`

= 0.0450 moles 

 

The limiting reagent is KOH so we will use the moles of Ni(OH)2 derived form KOH. We subtract the remaining 

`2 KOH (aq) + NiSO_4 (aq) -> Ni(OH)_2 (s) + K_2SO_4 (aq)`

    0.02124                0.0450                

  -0.01062                0.01062              0.01062         0.01062
_____________________________________________________

   0.01062 moles    0.03438 moles

Total volume = 118 + 300 = 418mL/1000 = 0.418L

Moles `Ni^2+` remain = moles NiSO4 remaining = 0.03438 moles

concentration of `Ni^(2+)` remaining = `0.03438/0.418` = 0.0822 M

concentration of `SO_4^(2-)` = 0.0450/0.418 =  0.108 M

concentration of `K^+` = `0.02124/0.418` = 0.508 M

Sources:

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question