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Number of moles of `Ba(NO_3)2 " in " 25 cm^(3)` <br> ` " " = ( 0.05)/( 1000) xx 25.0 = 1.25 xx 10 ^(-3)` <br> Number of moles of NaF in 25.0 ` cm ^(3)` <br> ` " " = (0.02)/(1000) xx 25.0 = 5.0 xx 10 ^(-4)` <br> The total volume of the solution on mixing ` Ba(NO_3) _2` and NaF solutions <br> ` " " = 25.0 + 25.0= 50.0 cm ^(3)` <br> 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)` <br> ` therefore ` The concentration of `Ba(NO_3)_2` in solution <br> ` " " = (1.25 xx 10 ^(-3))/(50) xx 1000` <br> ` " " = 0.025 " mol " L^(-1)` <br> and , the concentration of NaF in solutions <br> ` " " = ( 5.0 xx 10 ^(-4))/( 50 ) xx 1000` <br> ` " " = 0.01 " mol " L^(-1)` <br> Hence , ` [Ba^(2+)] = [Ba(NO_3) _2]` in solution <br> ` " " =0.025 xx (0.01)^(2) = 2.5 xx 10^(-6)` <br> 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.