Verified Solution • 2025–26 Session moderate

QJEEM09Ch3202: JEE Main 2009 Chemistry - Class 11 JEEM09Ch3202 Solution: Solid Ba(NO 3 ) 2 is gradually dissolved in a 1.0 10 -4 M Na 2…

Solid Ba(NO3)2 is gradually dissolved in a 1.0 ×\times 10-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form ?
(Ksp for BaCO3 = 5.1 ×\times 10−9 )

Multiple Choice Options

  • A 5.1 ×\times 10-5 M
  • B 8.1 ×\times 10-8 M
  • C 8.1 ×\times 10-7 M
  • D 4.1 ×\times 10-5 M

Step-by-Step Verified Solution

Na2CO31×10−4M→2Na+1×10−4M  +  CO32−1×10−4M\mathop {N{a_2}C{O_3}}\limits_{1 \times {{10}^{ - 4}}M} \to \mathop {2N{a^ + }}\limits_{1 \times {{10}^{ - 4}}M} \,\, + \,\,\mathop {C{O_3}^{2 - }}\limits_{1 \times {{10}^{ - 4}}M}

KSP(BaCO3)=[Ba2+][CO32−]{K_{SP\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right]

[Ba2+]=5.1×10−91×10−4\left[ {B{a^{2 + }}} \right] = {{5.1 \times {{10}^{ - 9}}} \over {1 \times {{10}^{ - 4}}}}

=5.1×10−5M= 5.1 \times {10^{ - 5}}M

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