x mg of Mg(OH)$_2$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer) (Given: Mg(OH)$_2$ is assumed to dissociate completely in H$_2$O)
x mg of Mg(OH)$_2$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer) (Given: Mg(OH)$_2$ is assumed to dissociate completely in H$_2$O)
Show Hint
Correct Answer: 3
Approach Solution - 1
Solution:
To find the amount of Mg(OH)2 needed to produce a pH of 10.0, we first determine the corresponding hydroxide ion concentration [OH−].
1. Calculate pOH from the given pH:
pOH = 14.0 - 10.0 = 4.0
2. Calculate [OH−]:
[OH−] = 10−pOH = 10−4.0 = 1.0 × 10−4 M
3. Because Mg(OH)2 dissociates completely in water as follows:
Mg(OH)2(s) → Mg2+(aq) + 2OH−(aq)
The concentration of OH− is twice the molarity of Mg(OH)2 in solution:
Let the concentration of Mg(OH)2 = [Mg(OH)2].
Thus, 2[Mg(OH)2] = [OH−]
[Mg(OH)2] = [OH−] / 2 = 0.5 × 10−4 M
4. Calculate the amount (in moles) of Mg(OH)2 needed for 1.0 L solution:
Moles of Mg(OH)2 = [Mg(OH)2] × Volume (L) = 0.5 × 10−4 mol/L × 1.0 L = 0.5 × 10−4 mol
5. Convert moles to grams:
Molar mass of Mg(OH)2 = 58 g/mol
Mass = moles × molar mass = 0.5 × 10−4 mol × 58 g/mol = 2.90 × 10−3 g
6. Convert grams to milligrams:
Mass = 2.90 × 10−3 g × 1000 mg/g = 2.90 mg
Rounded to the nearest integer, x = 3 mg.
Thus, the value of x is 3 mg, which fits within the provided range (3,3).
Approach Solution -2
- pH = 10.0
- pOH = 4.0
- [OH$^-$] = $10^{-4}$ M
The concentration of OH$^-$ is $10^{-4}$ M, and since Mg(OH)$_2$ dissociates completely in water, the number of moles of OH$^-$ will be equal to the number of moles of Mg(OH)$_2$. Step-by-step solution:
1. Number of moles of OH$^-$ = $10^{-4}$ moles (from the concentration of OH$^-$).
2. Number of moles of Mg(OH)$_2$ = $\frac{10^{-4}}{2}$ = $5 \times 10^{-5}$ moles, because one mole of Mg(OH)$_2$ gives two moles of OH$^-$.
3. The mass of Mg(OH)$_2$ is then calculated as: \[ \text{mass of Mg(OH)$_2$} = 5 \times 10^{-5} \times 58 \times 10^3 \, \text{mg} \] \[ = 2.9 \, \text{mg} \]
Thus, the value of x is 2.9 mg, which is approximately 3 mg.
Top Questions on Equilibrium
- A vessel at 1000 K contains \( \text{CO}_2 \) with a pressure of 0.5 atm. Some of \( \text{CO}_2 \) is converted into \( \text{CO} \) on addition of graphite. If total pressure at equilibrium is 0.8 atm, then \( K_p \) is:
- JEE Main - 2025
- Chemistry
- Equilibrium
- In the reaction \( \mathrm{N_2(g)} + 3\mathrm{H_2(g)} \leftrightarrow 2\mathrm{NH_3(g)} \), if the equilibrium constant \( K_c = 4 \times 10^{-3} \) at a certain temperature, which of the following is true about the reaction at equilibrium?
- BITSAT - 2025
- Chemistry
- Equilibrium
- In which of the following \( K_c = K_p \)? a) \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \) b) \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \)
- KEAM - 2025
- Chemistry
- Equilibrium
- For the reaction \( \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightleftharpoons 2\text{NH}_3 (\text{g}) \), the equilibrium constant \( K_c \) is \( 4.0 \times 10^{-2} \) at a certain temperature. If the equilibrium concentrations are \( [\text{N}_2] = 0.5 \, \text{M} \) and \( [\text{H}_2] = 1.5 \, \text{M} \), what is the equilibrium concentration of \( \text{NH}_3 \)?
- MHT CET - 2025
- Chemistry
- Equilibrium
- For the equilibrium reaction: \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] At a certain temperature, the equilibrium constant \( K_c \) is 4.0. If the concentrations of \( N_2 \), \( H_2 \), and \( NH_3 \) are 0.2 M, 0.6 M, and 0.4 M, respectively, calculate the reaction quotient \( Q_c \) and determine whether the reaction is at equilibrium.
- MHT CET - 2025
- Chemistry
- Equilibrium
Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
If \( S \) and \( S' \) are the foci of the ellipse \[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \] and \( P \) is a point on the ellipse, then \[ \min (SP \cdot S'P) + \max (SP \cdot S'P) \] is equal to:
- JEE Main - 2025
- Conic sections
- The center of mass of a thin rectangular plate (fig - x) with sides of length \( a \) and \( b \), whose mass per unit area (\( \sigma \)) varies as \( \sigma = \sigma_0 \frac{x}{ab} \) (where \( \sigma_0 \) is a constant), would be
- JEE Main - 2025
- System of Particles & Rotational Motion
- Let ABC be a triangle formed by the lines \( 7x - 6y + 3 = 0 \), \( x + 2y - 31 = 0 \), and \( 9x - 2y - 19 = 0 \). Let the point \( (h, k) \) be the image of the centroid of \( \triangle ABC \) in the line \( 3x + 6y - 53 = 0 \). Then \( h^2 + k^2 + hk \) is equal to:
- JEE Main - 2025
- Coordinate Geometry
Given below are two statements I and II.
Statement I: Dumas method is used for estimation of "Nitrogen" in an organic compound.
Statement II: Dumas method involves the formation of ammonium sulfate by heating the organic compound with concentrated H\(_2\)SO\(_4\). In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Nitrogen