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If 25.0 cm^(3) of 0.05 M Ba(NO(3))(2) so...

If `25.0 cm^(3)` of 0.05 M `Ba(NO_(3))_(2)` solution is mixed with 25.0 `cm^(3)` of 0.02 M NaF solution. Will any `BaF_(2)` precipitated ? `(K_(sp) " for " BaF_(2) =1.7 xx 10^(-6))`

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To determine if any BaF₂ will precipitate when mixing 25.0 cm³ of 0.05 M Ba(NO₃)₂ with 25.0 cm³ of 0.02 M NaF, we will follow these steps: ### Step 1: Calculate the moles of Ba²⁺ and F⁻ in the solutions. 1. **Calculate moles of Ba²⁺ from Ba(NO₃)₂:** - Volume of Ba(NO₃)₂ solution = 25.0 cm³ = 0.025 L - Concentration of Ba(NO₃)₂ = 0.05 M - Moles of Ba²⁺ = Volume (L) × Concentration (M) ...
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