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 × 10⁻⁶ at 298 ~K. The ratio of [Ba²⁺][F⁻]² and K sp is _ _ _ _ _ _. (Nearest integer)

Equilibrium: JEE Main 2023 Chemistry question

JEE Main 2023 · 13 Apr, Morning shift · Q11

JEE Main 202313 Apr, Morning shiftChemistryHardEquilibrium123× since 2002

25.0 \mathrm{~mL}

0.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}

is mixed with

25.0 \mathrm{~mL}

0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}}

\mathrm{BaF}_{2}

0.5 \times 10^{-6}

298 \mathrm{~K}

. The ratio of

\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}

and

\mathrm{K}_{\mathrm{sp}}

is ___________. (Nearest integer)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2023 Equilibrium Question with Answer (13 Apr Morning shift, Q11) | JEE Arc