H3PO4 + NO2- e HNO3 + H2PO4
1 answer:
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Answer: -
12.59
Explanation: -
Strength of NaOH = 0.0179 M
Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L
Number of moles = 0.0179 M x 0.058 L
= 1.04 x 10⁻³ mol
Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol
Strength of Ba(OH)₂ = 0.0294 M
Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L
Number of moles = 0.0294 M x 0.060 L
= 1.76 x 10⁻³ mol
Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol
Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol
= 4.56 x 10⁻³ mol
Total volume of the mixture = 58.0 + 60.0
= 118.0 mL
118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]
1000 mL of the solution has
= 0.0386 mol
Using the relation
pOH = - log [OH-]
= - log 0.0386
= 1.41
Using the relation
pH + pOH = 14
pH = 14 - 1.41
= 12.59