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An analytical chemist is titrating 241.8mL of a 1.200M solution of nitrous acid HNO2 with a 0.5200M solution of NaOH. Th...

An analytical chemist is titrating 241.8mL of a 1.200M solution of nitrous acid HNO2 with a 0.5200M solution of NaOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 605.5mL of the NaOH solution to it.

Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added.

Round your answer to 2 decimal places.

An analytical chemist is titrating 241.8 mL of a 1.200 M solution of nitrous acid (HNO2) with a 0.5200 M solution of NaOH. Th

An analytical chemist is titrating 241.8 mL of a 1.200 M solution of nitrous acid (HNO2) with a 0.5200 M solution of NaOH. The pK, of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 605.5 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. p
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Answer #1

use:

pKa = -log Ka

3.35 = -log Ka

Ka = 4.467*10^-4

Given:

M(HNO2) = 1.2 M

V(HNO2) = 241.8 mL

M(NaOH) = 0.52 M

V(NaOH) = 605.5 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 1.2 M * 241.8 mL = 290.16 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.52 M * 605.5 mL = 314.86 mmol

We have:

mol(HNO2) = 290.16 mmol

mol(NaOH) = 314.86 mmol

290.16 mmol of both will react

excess NaOH remaining = 24.7 mmol

Volume of Solution = 241.8 + 605.5 = 847.3 mL

[OH-] = 24.7 mmol/847.3 mL = 0.0292 M

use:

pOH = -log [OH-]

= -log (2.915*10^-2)

= 1.5353

use:

PH = 14 - pOH

= 14 - 1.5353

= 12.4647

Answer: 12.46

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