Respuesta :
The ph of the equivalence point = 7.851 and the volume of Koh needed to reach the equivalence point = 1.898 ml
Volume of HNO2 = V1 = 23.4 ml
Concentration of HNO2 = M1 = 0.0390 m
Concentration of koh = M2 = 0.0574 m
Volume of koh = V2
M1V1 = M2V2
(0.0390) (23.4) = (0.0574) V2
⇒V2 = 15.898 ml
The volume of KOH needed to reach equivalence point V2 = 15.898 ml
We need
total volume of solution = volume of hno2 + volume of KOH = 23.4 + 15.898 = 39.298 ml = 0.0393 L
the reaction we want is this:
NO2- + H2O <--> HNO2 + OH-
Number of moles of NO2- = molarity of HNO2 * volume of HNO2 = 0.0390 * 23.4 * (1L/10^3 ml) = 9.13 * 10 ^ (-4) mol
Concentration of NO2- at equivalence point = 9.13 * 10^(-4)/ 0.0393 = 0.0232 M
Kb = [HNO2][OH-] / [NO2-]
⇒2.17 * 10^(-11) = x * x / 0.0232
⇒x^2 = 5.0344 * 10^(-13)
⇒x = [OH-] = 7.09 * 10^(-7)
pOH = -log[OH-] = -log(7.09 * 10^(-7) ) = 6.149
pH = 14 - pOH = 14 - 6.149 = 7.851
Thus the pH of the equivalence point = 7.851
Find the pH of a solution and learn more about it
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