Question

0.74g ca(oh)2 is dissolved in water to give 200ml solution at 298 k. The hydrogen ion concentration in the solution is

Answer 1

here is your answer hope it help.

>>Mass of Ca(OH)2 =0.74g

-- Molar mass of Ca(OH)2 = 74g/mol.

Number of mole of Ca(OH)2 = 0.74/74 = 0.01

now, we find molarity..... ⤵️

➡️Molarity = ⤵️

no. of moles of Ca(OH)2/vol.of solution in L.

= (0.01/200)×1000 = 0.05M.

➡️ Ca(OH)2 dissociate.....

>> Ca(OH)2 ↔️ Ca^2+ + 2OH^-1

so,, conc. of ca(OH) 2 = 2× conc. of [OH-].
= 2× 0.05 = 0.1M

>> 10^-14 = [H+] [OH-]

>>>[H +] = 10^-14 / 0.1 = 10^-13 M.

✴️✴️ therefore 10^-13 M is the hydrogen ion concentration in the solution.

_________✌️

Answer 2

The hydrogen ion concentration in the solution is 10⁻¹³ M.

Given:

Mass of Ca(OH)₂ = 0.74 g

Volume of solution = 200 ml

Explanation:

The dissociation of Ca(OH)₂ is:

Ca(OH)₂ ↔ Ca²⁺ + 2 OH⁻¹

The molar mass of Ca(OH)₂ is:

Ca(OH)₂ = 40 + ((16 + 1) × 2) = 74 g/mol

∴ Number of moles = 0.74/74 = 0.01 moles

Molarity = (Number of moles)/(Volume of solution)

Molarity = (0.01/200) × 1000

∴ Molarity = 0.05 M

Concentration of calcium hydroxide:

[Ca(OH)₂] = 2 × [(OH)₂]

∴ [Ca(OH)₂] = 0.05 × 2 = 0.1 M

Equilibrium Concentration is given as:

[H⁺][OH⁻] = 10⁻¹⁴

[H⁺] = 10⁻¹⁴/[OH⁻]

[H⁺] = 10⁻¹⁴/0.1

∴ [H⁺] = 10⁻¹³ M

@gAvYa