Does a precipitate form when the aqueous solutions #"Ba(NO"_3)_2"# and #"2KOH"# are combined?

No, a precipitate will not form when barium nitrate, #Ba(NO_3)_2#, reacts with potassium hydroxide, #KOH#, in aqueous solution because both products are soluble in water.

The balanced chemical equation looks like this

#Ba(NO_3)_(2(aq)) + 2KOH(aq) -> Ba(OH)_2(aq) + 2KNO_3(aq)#

You are essentially dealing with a double replacement reaction; in aqueous solution, both reactants will dissociate into ions.

The complete ionic equation is

#Ba_((aq))^(2+) + 2NO_(3(aq))^(-) + 2K_((aq))^(+) + 2OH_((aq))^(-) -> Ba_((aq))^(2+) + 2OH_((aq))^(-) + 2K_((aq))^(+) + 2NO_(3(aq))^(-)#

Notice that all the ions that are on the reactants' side can also be found on the products' side. This implies that all the species will exist as ions in solution and no precipitate will be formed.

A quicker way to determine whether or not a reaction will produce a precipitate is to look at the solubility rules. If you're familiar with these rules, you'll notice that both barium hydroxide, #Ba(OH)_2#, and potassium nitrate, #KNO_3#, are soluble in aquesous solution.

The solubility rules:

http://www.csudh.edu/oliver/chemdata/solrules.htm

A precipitation reaction is a type of double replacement reaction, in which the cations and anions switch partners, and one of the products is a precipitate.

The evidence that a precipitation reaction has occurred is the formation of an insoluble solid (the precipitate) when two aqueous solutions are combined.

#"Ba(NO"_3)_2("aq")# + #"2KOH(aq)"# #rarr# #"Ba(OH)"_2"# + #"2KNO"_3"#

So how can we know whether #"Ba(OH)"_2"# or #"KNO"_3# is a precipitate? We need to consult a table of solubility rules.

Notice that all nitrates #("NO"_3^(-)")# are soluble, so #"KNO"_3# is soluble, and hydroxides #("OH"^(-)")# tend to be insoluble, but combined with #"Ba"^(2+)# is an exception, so #"Ba(OH)"_2"# is also soluble. So this reaction does not take place. What you have is a mixture of aqueous ions:

#"Ba"^(2+)("aq") + "2NO"_3^(-)("aq") + "2K"^(+)("aq") + "2OH"^(-)("aq")"##rarr##"Ba"^(2+)("aq") + "2OH"^(-)("aq") + "2K"^(+)("aq") + "2NO"_3^(-)("aq")"#

So you would write the equation as:

#"Ba(NO"_3)_2("aq")# + #"KOH(aq)"##rarr##"no reaction"#