Question

Be sure to answer all parts.

Find the pH of the equivalence point and the volume (mL) of 0.0822 M KOH needed to reach the equivalence point in the titration of 23.4 mL of 0.0390 M HNO₂.

Volume:

mL KOH

pH =

Answer 1

KOH + HNO KNO₃ + H₂O One mole KOH reacts with one mole HNO. So, (KOH) (ENG) MV - M₂ V2 (0.0022M) x V, - (0.0390 m) x (23.4 mL) Vi= (0.0390 m) x (23.4 mL) 0.0022m) VE PARS IV. = 11.1 mL) So, volume of Kot required = 11.1 mL Now, Moles of KOH = Molarity x Volume = (0.08 22M) (11.0mL) - 0.912 mole Moles of HNO₃ = Molarity X Volume = (0.0390m) (23.4mL) = 0.913 mole Mole of HNO > Mole of KOH 0.912 mole of HNO is neutralised by 0.912 mole of koH.

emaning Mole NO2 Remaining mole - 0.913 - of HNO2 0.912 = 0.001 mole. Total volume of solution - 23.4 mL + 11.1 mL = 34.5 mL -0.03452 Molarity of [+] = = 0.0289 m 0.0ol mol 0.0345L pH = -log Trt] = -log (0.0209) - - (-1.54) = 1.54 Hence, volume of KOH= 11.1 mL pH = 1.54.