Question

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

Answer 1

given, 0.74g Ca(OH)2 is dissolved in water to give 200mL solution at 298K.

mass of Ca(OH)2 = 0.74g
molar mass of Ca(OH)2 = 40 + (16 + 1) × 2
= 40 + 34 = 74g/mol

so, number of moleof Ca(OH)2 = 0.74/74 =0.01

now, molarity = number of moles of Ca(OH)2/volume of solution in L

= {0.01/200} × 1000 = 0.05M

Ca(OH)2 dissociate , $Ca(OH)_2\leftrightarrow Ca^{2+}+2OH^{-1}$

so, concentration of Ca(OH)2 = 2 × concentration of [OH-]
= 0.05 × 2 = 0.1M

we know , $10^{-14}=[H^+][OH^-]$
$[H^+]$ = 10^-14/0.1 = 10^-13 M

hence, concentration of hydrogen ion is 10^-13M

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