If ` 25.cm ^(3) " of " 0.050 " M " Ba(NO_3) _2` are mixed with ` 25.0 cm ^(3)` of 0.020 M NaF, will any `BaF_2 ` precipitated `K_(sp) " of " BaF_2 " is " 1.7 xx 10 ^(-6) " at " 298K`.

Explanation

Number of moles of `Ba(NO_3)2 " in " 25 cm^(3)`
` " " = ( 0.05)/( 1000) xx 25.0 = 1.25 xx 10 ^(-3)`
Number of moles of NaF in 25.0 ` cm ^(3)`
` " " = (0.02)/(1000) xx 25.0 = 5.0 xx 10 ^(-4)`
The total volume of the solution on mixing ` Ba(NO_3) _2` and NaF solutions
` " " = 25.0 + 25.0= 50.0 cm ^(3)`
This means that ` 1.25 xx 10 ^(-3)` moles of `Ba(NO_3) and 5.0 xx 10 ^(-4)` moles of NaF are now present in ` 50 cm ^(3)`
` therefore ` The concentration of `Ba(NO_3)_2` in solution
` " " = (1.25 xx 10 ^(-3))/(50) xx 1000`
` " " = 0.025 " mol " L^(-1)`
and , the concentration of NaF in solutions
` " " = ( 5.0 xx 10 ^(-4))/( 50 ) xx 1000`
` " " = 0.01 " mol " L^(-1)`
Hence , ` [Ba^(2+)] = [Ba(NO_3) _2]` in solution
` " " =0.025 xx (0.01)^(2) = 2.5 xx 10^(-6)`
The value of `K_(sp) " for " BaF_2 " is " 1.7 xx 10 ^(-6) ` Since the value of ionic product is greater than that of the solubility product , precipitation of `BaF_2 ` will occur.