Question

A solution is prepared by dissolving 2:45g of H2SO4 in 500ml of solution. Calculate pH of th solution?​

Answer 1

Answer:

The pH of the solution is 1.

Explanation:

Given mass of $H_{2}SO_{4}$ = 2.45 g

Molar mass of $H_{2}SO_{4}$ = (2 × 1) + (1 × 32) + (4 × 16) = 98 g/mol

Number of moles of $H_{2}SO_{4}$ = given mass ÷ molar mass

                                              = $\frac{2.45}{98}$ = 0.025 moles

Given volume of the solution = 500 mL = 0.5 L

We know that molarity = number of moles of solute ÷ volume of solution in litres.

= $\frac{0.025}{0.5}$

= 0.05 M

⇒Concentration of the given solution = 0.05 M

$H_{2}SO_{4}$ ⇄ $2H^{+} + SO_{4} ^{2-}$

∴Concentration of $H^{+}$ = 0.05 × 2 = 0.1 M

We know that pH = -log[$H^{+}$]

                             = -log(0.1)

                             = 1

Therefore, the pH of the solution is 1.

Answer 2

Answer:

the pH of H₂SO₄ solution (2.45g in 500mL) is 1.

Explanation:

Given,

        mass of H₂SO₄ = 2.45g

   volume of solution = 500mL = 0.5L

molar mass of H₂SO₄ = 98g/mol.

                H₂SO₄ → 2H⁺ + SO₄²⁻

1 mole of H₂SO₄ gives 2 moles of H⁺ ion.

  moles of H₂SO₄ = mass of H₂SO₄ / molar mass of H₂SO₄

⇒                           = 2.45 / 98 = 0.025moles

molar concentration = moles of H₂SO₄  / liters of solution

⇒                                = 0.025/ .5 = 0.05M

then the concentration of H⁺ is,

               [H⁺] = 2 × 0.05 = 0.1M

pH of the solution is given by,

       pH= - log [H⁺]

⇒ = - log(0.1) = 1

Hence, the required answer is 1.