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Chemistry Question 71 – JEE-MAIN 2025

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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 H2O)

Recall that at 298 K, the sum of pH and pOH is 14.

Step 1: Determine Hydroxide Ion Concentration✦ Active

Given pH = 10.0. At 298 K, pH+pOH=14. Calculate pOH:

pOH=14−pH=14−10.0=4.0

Now, calculate the hydroxide ion concentration [OH−]:

[OH−]=10−pOH=10−4.0M
Step 2: Calculate Mg(OH)2 Molarity○ Expand

Magnesium hydroxide dissociates completely as follows:

Mg(OH)2(s)→Mg2+(aq)+2OH−(aq)

From the stoichiometry, [OH−]=2×[Mg(OH)2]. Therefore, the molarity of Mg(OH)2 required is:

[Mg(OH)2]=[OH−]2=10−4M2=0.5×10−4M=5×10−5M
💡 Teacher's Secret Hint

Ensure to account for the stoichiometry of OH− ions produced per mole of Mg(OH)2.

Step 3: Calculate Mass of Mg(OH)2○ Expand

The volume of the solution is 1.0 L. Molar mass of Mg(OH)2 is 58 g/mol. Calculate the moles of Mg(OH)2:

Moles of Mg(OH)2=Molarity×Volume=(5×10−5 mol/L)×(1.0 L)=5×10−5 mol

Now, calculate the mass in grams:

Mass of Mg(OH)2=Moles×Molar Mass=(5×10−5 mol)×(58 g/mol)=290×10−5 g=2.9×10−3 g

Convert the mass to milligrams:

x=2.9×10−3 g×1000 mg/g=2.9 mg

The value of x to the nearest integer is 3.

💡 Teacher's Secret Hint

Pay attention to unit conversions, especially from grams to milligrams.

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