[navneet]

Hi

Today we did a experiment, where we mixed 2.70 grams of potassium hydroxide solid (KOH), into distelled water. Can you please tell me if my equation is blanced correct and  if this is the proper equation, and what would KOOH be (solid, gas, liquid or aq), please correct me if im wrong.

The eqation I got:  2KOH_(s) + 2H₂O_(l) ==> 2KOOH + 2H_{2 (g)}

[tonyliruhan]

maybe your equation is incurrect.
are you sure there is some gas come out?
can you give us some details about the experiment

[navneet]

well its a dissolving experiment, we add solid KOH into H2O liquid and record the temp, and the temperature goes up, which means its a exothermic respect to KOH.  But Im not sure if thats how you write the equation for dissolving.

[mike]

KOH_(s) -----> K⁺_(aq)  +  OH^-_(aq)

[tonyliruhan]

I think there is no gas formed, the gas came out may be H2O
or you can mark this K is more active than Na, maybe the solid of
KOH is a kind of mixture of KOH and some oxidation of K which will
react with water and produce some gas

[navneet]

thank you im sure that is correct.

This is the experiment brifely

Reaction 1: Solid Potassium hydroxide dissolves in water to form an aqueous solution of ions.

Reaction 2: An aqueous solution of Potassium hydroxide reacts with an aqueous solution of hydrochloric acid.

Reaction 3: Solid Potassium hydroxide reactions with an aqueous solution of hydrochloric acid.

Reaction 1 equation: KOH_(s)==> K⁺ _(aq)+OH^- _(aq)

Reaction 2 equation: H⁺+OH^- _(aq)==> H₂O_(l)

Reaction 3 equation: KOH_(s)+H⁺==>H₂O_(l)+K⁺

after i get this it says in a question add the first net equation and the third and compare 2nd then, however it doesnt make sense because the equation is toally different when you add the 1st and 3rd, they are no where close to the 2nd equation, so is this question a trick question?

[Moonshyne]

KOH is base, so mixing it in water makes a basic solution that is in equilibrium. The equation should read:

KOH_(s) + H₂O_(l)<--> K⁺_(aq) + OH^-_(aq) + H₂O_(l)

There shouldn't be any chemical reaction, the KOH is just dissolving in the water. I am not sure how this can be exothermic...

[navneet]

Well for
reaction 1: temp rises by 3 degree C (from room temp (20 to 23)

reaction 2: temp stays the same

reaction 3: temp rises by 6 degree C (from room temp (20 to 26)

[AWK]

KOH, both in solid and in solution is in ionized form

[Borek]

KOH is base, so mixing it in water makes a basic solution that is in equilibrium. The equation should read:

KOH_(s) + H₂O_(l)<--> K⁺_(aq) + OH^-_(aq) + H₂O_(l)

Water cancels out.

There shouldn't be any chemical reaction, the KOH is just dissolving in the water. I am not sure how this can be exothermic...

Solvation is an exothermic process.

[Moonshyne]

Yeah I figured why... Even though the breaking of the bond between the K and the OH is an endothermic process, the formation of salvation shells around K⁺ and OH^- not only cancel the endothermicity, but make the total process exothermic. Perticularly the shell around the OH^- contributes most of the heat to the process.

I know water cancels out in the equation I previously wrote--I was just trying to show what happens when KOH is added to water, as requested.