Chemistry, asked by aadhiii68, 4 months ago

50 ml of 0.001N HCl and 10ml of 0.01N H2SO4 are mixed together. Calculate pH of resulting solution?​

Answers

Answered by malavikathilak123
8

Answer:

pH of resulting solution of50 ml of 0.001N HCl and 10ml of 0.01N H₂SO₄ is 2.6.

Explanation:

Given that,

        volume of HCl = 50mL= 0.05L

     normality of HCl = 0.001N

   volume of H₂SO₄ = 10mL = 0.01L

normality of  H₂SO₄ = 0.01N

molarity of  HCl = 0.001M

molarity of H₂SO₄ = 0.01/2=0.005M

number of moles = molarity × liters of solution.

1 mole of HCl gives 1 mole of  H⁺ and 1 mole of H₂SO₄ of gives 2 moles of  H⁺

     moles of H⁺ in HCl = 0.001M×0.05L = 5×10⁻⁵mol.

moles of H⁺ in H₂SO₄ = 2×0.005×0.01 = 1×10⁻⁴=10×10⁻⁵mol.

       total moles of H⁺ = 5×10⁻⁵ + 10×10⁻⁵ = 15×10⁻⁵ =1.5×10⁻⁴mol.

    volume of solution = 0.05+0.01 = 0.06L

total concentration of H⁺ is ,

                            [H⁺] = 1.5×10⁻⁴mol /  0.06L = 2.5×10⁻³ M

pH of the solution is given by,

             pH = -log[H⁺]

⇒                 = -log 2.5×10⁻³ = 2.6

the pH of resulting solution is 2.6.

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