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When a 25.9 mL sample of a 0.317 M aqueous nitrous acid solution is titrated with a 0.458 M aqueous sodium hydroxide solution, what is the pH after 26.9 mL of sodium hydroxide have been added? Please show work!

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Answer #1

millimoles of HNO2 = 25.9 x 0.317 = 8.21

millimoles of NaOH = 0.458 x 26.9 = 12.32

Ka of HNO2 = 5.6×10–4

pKa of HNO2 = 3.25

HNO2 + NaOH   -----------> NaNO2 + H2O

8.21      12.32                            0             0

0          4.11                            8.21

here strong acid remains. so

[NaOH] = 4.11 / (25.9 + 26.9) = 0.0778

pOH = -log [OH-] = -log (0.0778)

poH = 1.11

pH = 12.89

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