Question

8 g naoh and 4.9g h2so4 are present in one litre of the solution . What is its ph

Answer 1

Answer:

The pH of the solution is 13

Explanation:

Molecular weight of NaOH = 40

Weight of NaOH = 8g

Moles of NaOH = Weight/Molecular weight

                         = 8/40 = 0.2 moles

Molecular weight of H₂SO₄ = 98

Weight of H₂SO₄ = 4.9g

Moles of H₂SO₄ = Weight/Molecular weight

                         = 4.9/98 = 0.05 moles

1 mole of NaOH will give = 1 mole of OH⁻ ions

0.2 moles of NaOH will give = 0.2 moles of OH⁻ ions

1 mole of H₂SO₄ will give = 2 moles of H⁺ ions

0.05 moles of H₂SO₄ will give = 0.05 × 2 = 0.1 moles of H⁺ ions

Thus 0.1 moles of H⁺ will be neutralised by 0.1 moles of OH⁻ ions

And 0.1 moles of OH⁻ ions will be left

Concentration of OH⁻ ions = 0.1/1 = 0.1 M

or, [OH⁻ ] = 0.1 M

pOH = -log[OH⁻ ] = -log(0.1) = -log10⁻¹ = 1

We know that

pH + pOH = 14

Thus, pH + 1 = 14

or, pH = 14 - 1 = 13

Therefore pH of the solution is 13