Download App

STD 11 - 6.2. Equilibrium - II (icon Equilibrium) — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6.2. Equilibrium - II (icon Equilibrium)4 Marks
MCQ
Solid $Ba(NO_3)_2$ is gradually dissolved in $1\times10^{-4}\, M\, Na_2CO_3$ solution. At what concentration of $Ba^{+2}$ the precipitation of $BaCO_3$ will start? (If $K_{sp}$ of $BaCO_3 = 5.1\times10^{-9}$)
  • A
    $4.1\times10^{-5}\, M$
  • ✓
    $5.1\times10^{-5}\, M$
  • C
    $1.6\times10^{-4}\, M$
  • D
    $8.1\times10^{-7}\, M$
✓

Answer

Correct option: B.
$5.1\times10^{-5}\, M$
b
$\left[\mathrm{Ba}^{+2}\right]\left(10^{-4}\right)=5.1 \times 10^{-9}$

$\boxed{\left[ {{\text{B}}{{\text{a}}^{ + 2}}} \right] = 5.1 \times {{10}^{ - 5}}{\mkern 1mu} {\text{M}}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If uncertainty in the position of an electron is zero, the uncertainty in its momentum would be
→
The correct order in which the $O -O$ bond length increases in the following is
→
In which compound peroxide bond is present
→
Which is correct
→
Which of the following reactions is reversible
→
Which is represented by the formula ${C_n}{H_{2n - 2}}$
→
For hydrogen like atoms, the average distance of electron from nucleus for the orbitals belonging to the $4^{th}$ orbit, is minimum for
→
$p$-Amino- $N, N$-dimethylaniline is added to a strongly acidic solution of ${X}$. The resulting solution is treated with a few drops of aqueous solution of ${Y}$ to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of ${Y}$ with the reagent potassium hexacyanoferrate$(II)$ leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of ${Y}$ with the solution of potassium hexacyanoferrate$(III)$ leads to a brown coloration due to the formation of ${Z}$.

$1.$ The compound ${X}$ is

$(A)$ $\mathrm{NaNO}_3$ $(B)$ $\mathrm{NaCl}$ $(C)$ $\mathrm{Na}_2 \mathrm{SO}_4$ $(D)$ $\mathrm{Na}_2 \mathrm{~S}$

$2.$ The compound ${Y}$ is

$(A)$ $\mathrm{MgCl}_2$ $(B)$ $\mathrm{FeCl}_2$ $(C)$ $\mathrm{FeCl}_3$ $(D)$ $\mathrm{ZnCl}_2$

$3.$ The compound ${Z}$ is

$(A)$ $\left.\mathrm{Mg}_2 \mid \mathrm{Fe}(\mathrm{CN})_6\right]$ $(B)$ $\operatorname{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(D)$ $\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

Give the answer question $1,2$ and $3.$

→
Identify $A$ and predict the type of reaction.

 $\xrightarrow{{NaN{H_2}}}\,A$

→
The $pH$ of water at ${25\,^o}C$ is nearly
→