Question

25.0 mL of 0.050 M Ba(NO₃)₂ is mixed with 25.0 mL of 0.020 M NaF. K_sp of BaF₂ is 0.5 x 10^-6 at 298 K. The ratio of [Ba2+] [F^-]² and K_sp is__ (Nearest integer)

Answer 1

Correct Answer: 5

We are given the following data:

Volume of Ba(NO₃)₂ = 25.0 mL,

Concentration of Ba(NO₃)₂ = 0.050 M,

Volume of NaF = 25.0 mL,

Concentration of NaF = 0.020 M,

\( K_{sp} \) of BaF₂ = \( 0.5 \times 10^{-6} \).

First, calculate the concentrations of Ba²⁺ and F^- after mixing. The total volume after mixing is: \[ V_{\text{total}} = 25.0 \, \text{mL} + 25.0 \, \text{mL} = 50.0 \, \text{mL}. \] Now, the concentration of Ba²⁺ after mixing: \[ [\text{Ba}^{2+}] = \frac{0.050 \times 25.0}{50.0} = 0.025 \, \text{M}. \] The concentration of F^- after mixing: \[ [\text{F}^{-}] = \frac{0.020 \times 25.0}{50.0} = 0.010 \, \text{M}. \] The solubility product for BaF₂ is: \[ K_{sp} = [\text{Ba}^{2+}][\text{F}^{-}]^2. \] Substitute the values: \[ K_{sp} = (0.025) \times (0.010)^2 = 0.5 \times 10^{-6}. \] We are asked for the ratio of \([\text{Ba}^{2+}]\) to \(K_{sp}\). This is given by: \[ \frac{[\text{Ba}^{2+}]}{K_{sp}} = \frac{0.025}{0.5 \times 10^{-6}} = 5 \times 10^4. \] Thus, the ratio is \( 5 \times 10^4 \).