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Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 100.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 × 10⁻⁶.

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Answer:

pH = 1.85

Explanation:

The reaction of H₂NNH₂ with HNO₃ is::

H₂NNH₂ + HNO₃ → H₂NNH₃⁺ + NO₃⁻

Moles of H₂NNH₂ and HNO₃ are:

H₂NNH₂: 0.0400L ₓ (0.200mol / L) = 8.00x10⁻³ moles of H₂NNH₂

HNO₃: 0.1000L ₓ (0.100mol / L) = 0.01 moles of HNO₃

As moles of HNO₃ > moles of H₂NNH₂, all H₂NNH₂ will react producing H₂NNH₃⁺, but you will have an excess of HNO₃ (Strong acid).

Moles of HNO₃ in excess are:

0.01 mol - 8.00x10⁻³ moles = 2.00x10⁻³ moles of HNO₃ = moles of H⁺

Total volume is 100.0mL + 40.0mL = 140.0mL = 0.1400L.

Thus, [H⁺] is:

[H⁺] = 2.00x10⁻³ moles / 0.1400L = 0.0143M

As pH = - log [H⁺]

pH = 1.85

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The pH of the resulting solution is 1.85.

The reaction is 1:1 hence;

Number of moles of  H₂NNH₂ = 40/1000 L × 0.200 M = 0.008 moles

Number of moles of  HNO₃ = 100/1000 L × 0.100 M = 0.01 moles of HNO3

Number of moles of excess acid = 0.01 moles - 0.008  = 0.002 moles

Total volume of solution;

40.0 mL + 100.0 mL = 140 mL or 0.14 L

Molarity of excess acid = 0.002 moles/ 0.14 L = 0.014 M

Since;

pH = -log[H^+]

pH = -log[0.014 M]

pH = 1.85

The pH of the resulting solution is 1.85.

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