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 JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Equilibrium Questions
- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- For the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \), \( K_p = 0.492 \, \text{atm} \) at 300 K. \( K_c \) for the reaction at the same temperature is _____ \( \times \, 10^{-2} \).
(Given: \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \)) - \( A_{(g)} \rightleftharpoons B_{(g)} + \frac{C}{2}_{(g)} \) The correct relationship between \( K_P \), \( \alpha \), and equilibrium pressure \( P \) is:
- The equilibrium constant for the reaction
\( \text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g) \)
is \( K_C = 4.9 \times 10^{-2} \). The value of \( K_C \) for the reaction given below is
\( 2 \text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{SO}_3 \, (g) \) is - 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:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,