Respuesta :
When a 29.2 ml sample of a 0.496 m aqueous nitrous acid solution is titration with a 0.449 m aqueous sodium hydroxide solution, The ph after 48.4 ml of sodium hydroxide have been added is 13.07.
Given Volume of HNO2 = 29.2 ml
concentration of HNO2 = 0.496 ml
No of molars of HNO2 = 29.2 × 0.486 = 14.1912 millimole
Given volume of NaOH = 48.4 ml
concentration of NaOH = 0.496 ml
No of moles of NaOH = 48.4 × 0.496 = 24.0064 millimole
Excess NaOH = 24.0064 - 14.1912 = 9.8152 millimoles
Total volume = 48.4 + 29.2 =77.6 ml
concentration of NaOH = 9.8152 / 77.6 =0.12648m
poh = -log[oH-] = -log(0.12) = 0.9208
ph = 14 - 0.9208 = 13.07
pka of HNO2 = 3.16
at mid point of titration [HNO2] = [NO2-]
ph = pka + log [NO2-]/[HNO2-]
= 3.16 + log 1
= 3.16
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