Chemistry question (1 Viewer)

[A.void.that.thinks]

Could somebody please help me with this question. The answer is D btw.

 

[CM_Tutor]

Barium nitrate is Ba(NO₃)₂ so a 0.0500 M solution has [Ba²⁺] = 0.0500 M and [NO₃^-] = 0.100 M.

Sodium hydroxide is NaOH so a 0.100 M solution has [Na⁺] = 0.100 M and [OH^-] = 0.100 M.

100 mL of each solution is combined to a 200 mL solution, so using the dilution formula C₁V₁ = C₂V₂ to find the diluted concentrations (C₂) from the starting concentrations (C₁) with V₁ = 100 mL and V₂ = 200 mL gives:

[Ba²⁺] = 0.0500 x 100 / 200 = 0.0250 M

and

[OH^-] = 0.100 x 100 / 200 = 0.0500 M.

Now, the solubility reaction that we are looking at is

Ba(NO₃)₂(s) <----> Ba²⁺(aq) + 2OH^-(aq)

and so

Q = [Ba²⁺][OH^-]² = 0.0250 x 0.0500² = 6.25 x 10^-5.

K_sp(Ba(OH)₂) = 2.55 x 10^-4 > Q

Since Q < K_sp, according to the dissolution equation, more barium hydroxide would need to dissolve to reach saturation, and so no precipitate will form.

 

[A.void.that.thinks]

Thank you!!!