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