.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $25.0 \mathrm{~mL}. Tags and topics: Equilibrium,Buffer Solution & Solubility,Degree of Dissociation, Le-chatelier, Reaction Quotient, G,JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions, Class 11 Chemistry,Class 1.">
.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $25.0 \mathrm{~mL}. Tags and topics: Equilibrium,Buffer Solution & Solubility,Degree of Dissociation, Le-chatelier, Reaction Quotient, G,JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions, Class 11 Chemistry,Class 1.">
.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $25.0 \mathrm{~mL}. Tags and topics: Equilibrium,Buffer Solution & Solubility,Degree of Dissociation, Le-chatelier, Reaction Quotient, G,JEE MAIN,JEE Advanced,JEE Mains,NCERT,NEET,Solutions, Class 11 Chemistry,Class 1.">
$25.0 \mathrm{~mL}$ of $0.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $25.0 \mathrm{~mL}$ of $0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}}$ of $\mathrm{BaF}_{2}$ is $0.5 \times 10^{-6}$ at $298 \mathrm{~K}$. The ratio of $\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}$ and $\mathrm{K}_{\mathrm{sp}}$ is ___________.
(Nearest integer)
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Step-by-Step Solution
Step 1: List the Known Quantities
• Volume of Ba(NO3)2 solution = 25.0 mL
• Concentration of Ba(NO3)2 solution = 0.050 M
• Volume of NaF solution = 25.0 mL
• Concentration of NaF solution = 0.020 M
• Ksp of BaF2 = 0.5 × 10−6
• Total volume after mixing = 25.0 mL + 25.0 mL = 50.0 mL
Step 2: Determine the New Concentrations After Mixing
After the solutions are mixed, each ion is diluted to a total volume of 50.0 mL.
For Ba2+ ions:
[\mathrm{Ba}^{2+}]
= \frac{\text{moles of Ba}^{2+}}{\text{total volume}}
= \frac{(25.0\ \mathrm{mL}) \times (0.050\ \mathrm{M})}{50.0\ \mathrm{mL}}
= 0.025\ \mathrm{M}.
For F− ions:
[\mathrm{F}^{-}]
= \frac{\text{moles of F}^{-}}{\text{total volume}}
= \frac{(25.0\ \mathrm{mL}) \times (0.020\ \mathrm{M})}{50.0\ \mathrm{mL}}
= 0.010\ \mathrm{M}.
Step 3: Calculate [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^{2}
Let Q be the ionic product, which has the same form as the solubility product:
Q = [\mathrm{Ba}^{2+}]\,[\mathrm{F}^{-}]^{2}.
Substitute the computed concentrations:
Q
= (0.025\ \mathrm{M}) \times (0.010\ \mathrm{M})^{2}
= 2.5 \times 10^{-6}.
Step 4: Find the Ratio of Q to K_{sp}
We are given:
K_{sp} (\mathrm{BaF}_{2}) = 0.5 \times 10^{-6}.
Hence,
\frac{Q}{K_{sp}}
= \frac{2.5 \times 10^{-6}}{0.5 \times 10^{-6}}
= 5.
Final Answer (Nearest Integer)
The ratio of [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^{2} to K_{sp} is 5.