Question

1/20 molar 500 ml H2SO4 solution is mixed with 1/10 molar 200 ml NaoH solution calaulate pH

Answer 1

Equivalents of H2SO4 = 2 × Moles
= 2 × Molarity × Volume
= 2 × 1/20 × 500 × 10^-3
= 0.05

Equivalents of NaOH = 1 × Moles
= 1 × Molarity × Volume
= 1 × 1/10 × 200 × 10^-3
= 0.02

Net equivalents = 0.05 - 0.02 = 0.03

Normality = Equivalents / Volume
= 0.03 / [(500 + 200) × 10^-3]
= 0.043 N

pH = - log[H+]
= - log 0.043
= 1.36

pH of resulting solution is 1.36 (acidic).

Answer 2

Answer:

Equivalents of H2SO4 = 2 × Moles

= 2 × Molarity × Volume

= 2 × 1/20 × 500 × 10^-3

= 0.05

Equivalents of NaOH = 1 × Moles

= 1 × Molarity × Volume

= 1 × 1/10 × 200 × 10^-3

= 0.02

Net equivalents = 0.05 - 0.02 = 0.03

Normality = Equivalents / Volume

= 0.03 / [(500 + 200) × 10^-3]

= 0.043 N

pH = - log[H+]

= - log 0.043

= 1.36

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