Verified Solution • 2025–26 Session moderate

QJEEM23Chb4bf: JEE Main 2023 Chemistry - Class 11 JEEM23Chb4bf Solution: of is mixed with of of is at . The ratio of and is ___________. …

25.0 mL25.0 \mathrm{~mL} of 0.050 M Ba(NO3)20.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} is mixed with 25.0 mL25.0 \mathrm{~mL} of 0.020 M NaF.KSp0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}} of BaF2\mathrm{BaF}_{2} is 0.5×10−60.5 \times 10^{-6} at 298 K298 \mathrm{~K} . The ratio of [Ba2+][F−]2\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2} and Ksp\mathrm{K}_{\mathrm{sp}} is ___________.

(Nearest integer)

Step-by-Step Verified Solution

Initial concentrations before mixing:

[Ba(NO3)2]=0.050 M[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}

[NaF]=0.020 M[\mathrm{NaF}] = 0.020\, \mathrm{M}

Volumes of the solutions:

VBa(NO3)2=VNaF=25.0 mLV_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}

After mixing, the total volume becomes:

Vtotal=25.0 mL+25.0 mL=50.0 mLV_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}

Now, we calculate the initial concentrations after mixing:

[Ba2+]=25.0 mL×0.050 M50.0 mL=0.025 M[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}

[F−]=25.0 mL×0.020 M50.0 mL=0.010 M[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}

Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂:

Q=[Ba2+][F−]2Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2

Q=(0.025 M)(0.010 M)2=2.5×10−6Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}

Ksp of BaF₂ is given as 5×10−75 \times 10^{-7} .

Now, we find the ratio of [Ba2+][F−]2[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2 to Ksp:

Ratio=(2.5×10−6)(5×10−7)=5\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5

So, the correct ratio of [Ba2+][F−]2[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2 to Ksp is 5.

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