Question #bf61e

1 Answer
Jul 22, 2017

Answer:

Yes, a precipitate will form. The procedure is the same as in your diagram.

Explanation:

Step 1. Determine the moles of each ion used

The equation for the reaction is:

#"NiF"_2"(aq)" + "2NaOH(aq)" → "Ni(OH)"_2"(s)" + "2NaF(aq)"#

#"Moles of OH"^"-"#

#= 0.0180 color(red)(cancel(color(black)("L NaOH solution"))) × (4.6 × 10^"-4" color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH solution")))) × "1 mol OH"^"-"/(1 color(red)(cancel(color(black)("mol NaOH")))) = 8.28 × 10^"-6" color(white)(l)"mol OH"^"-"#

#"Moles of Ni"^"2+"#

#= 0.0250 color(red)(cancel(color(black)("L NiF"_2 color(white)(l)"solution"))) × (8.92 × 10^"-4" color(red)(cancel(color(black)("mol NiF"_2))))/(1 color(red)(cancel(color(black)("L NiF"_2color(white)(l)color(white)(l) "solution")))) × "1 mol Ni"^"2+"/(1 color(red)(cancel(color(black)("mol NiF"_2))))#

#= 2.230 × 10^"-5" color(white)(l)"mol Ni"^"2+"#

Step 2. Determine the initial concentrations of each ion in the total volume

#V_text(tot) = "(0.0180 + 0.0250) L = 0.0430 L"#

#["Ni"^"2+"] = (2.230 × 10^"-5" color(white)(l)"mol")/("0.0430 L") = 5.186 × 10^"-4" "mol/L"#

#["OH"^"-"] = (8.28 × 10^"-6" color(white)(l)"mol")/("0.0430 L") = 1.926 × 10^"-4" "mol/L"#

Step 3. Compute the reaction quotient

#"Ni(OH)"_2 ⇌ "Ni"^"2+" + 2"OH"^"-"#

#Q = ["Ni"^"2+"]["OH"^"-"]^2 = 5.186 × 10^"-4" × (1.926 × 10^"-4")^2 = 1.922 × 10^"-11"#

Step 4. Compare #Q# to #K_text(sp)#

#Q = 1.922 × 10^"-11"#; #K_text(sp) = 5.48 × 10^"-16"#

#Q > K_text(sp)#, therefore #"Ni(OH)"_2# precipitates.