A 40.8 mL sample of a 0.582 M aqueous nitrous acid solution is titrated with a...

A 40.8 mL sample of a 0.582 M aqueous nitrous acid solution is titrated with a 0.285 M aqueous potassium hydroxide solution. What is the pH after 56.5 mL of base have been added?

Expert Solution

no of moles of HNO2 = molarity * volume in L

= 0.582*0.0408 = 0.0237456 moles

no of moles of KOH = molarity * volume in L

= 0.285*0.0565 = 0.0161025 moles

HNO2 (aq) + KOH(aq) ---------------> KNO2(aq) + H2O(l)

I 0.0237456 0.0161025 0

C -0.0161025 -0.0161025 0.0161025

E 0.0076431 0 0.0161025

Pka = 3.15

[KNO2] = 0.0161025mole

[HNO2] = 0.0076431moles

PH = PKa + log[KNO3]/[HNO2]

= 3.15 + log0.0161025/0.0076431

= 3.15 + 0.3236

= 3.4736>>>>>answer

queen_honey_blossom answered 2 years ago

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