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JEE Advanced 2025 Paper 2 Online
Numerical
+4
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The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $\boldsymbol{X} \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}$. The value of $\boldsymbol{X}$ is ____________.

Use: Solubility product constant $\left(K_{\mathrm{sp}}\right)$ of barium iodate $=1.58 \times 10^{-9}$

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2
JEE Advanced 2025 Paper 1 Online
Numerical
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At 25 °C, the concentration of H+ ions in 1.00 × 10−3 M aqueous solution of a weak monobasic acid having acid dissociation constant (Ka) of 4.00 × 10−11 is X × 10−7 M. The value of X is ______.

Use: Ionic product of water (Kw) = 1.00 × 10−14 at 25 °C

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3
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
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Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in a solution is $1 \mathrm{M}$ and $1.8 \times 10^{-2} \mathrm{M}$, respectively. Molar solubility of $\mathrm{PbSO}_{4}$ in the same solution is $\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}$ (expressed in scientific notation). The value of $Y$ is ________.

[Given: Solubility product of $\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

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4
JEE Advanced 2022 Paper 1 Online
Numerical
+3
-0
Change Language

A solution is prepared by mixing $0.01 \mathrm{~mol}$ each of $\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}$, and $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ of water. $p \mathrm{H}$ of the resulting solution is _________.

[Given: $p \mathrm{~K}_{\mathrm{a} 1}$ and $p \mathrm{~K}_{\mathrm{a} 2}$ of $\mathrm{H}_{2} \mathrm{CO}_{3}$ are $6.37$ and 10.32, respectively; $\log 2=0.30$ ]

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