Question

A 60.0 mL sample of 0.14 M NH₃ is titrated with 0.35 M HNO₃.

a. Write a balance equation between NH₃ and HNO₃.

b. Predict the pKa of NH₄+.

c. Determine the pH of the solution after the addition of 19.0 mL of HNO₃.

Answer 1

as Balanced equation between NH₂ and HNO base HNO₃ + H₂ NHA + No, acid base a conjugate conjugate 5 mg of NH3 = 18 810*5 * (the segna lace of boil) = 1044 > (ka of words) * (kb. 0f wins) – 16" >> Kas of within + vola n so, pka of wiha = -log (kan) . 10 Vil I Vog (5:55X10'0) -- js = 9.3 © FOM HNO3 0,00 19 m 0.35 M HNO-K for No. of moles of HNO3 = 0,35% 19 note ·motes - 1000 For NHB = 0.0066 moles 60 mL 0.14M NH3. No. of moles of NH3 = 0.14 x600.00 looo:

in Hill hul fitration it is seen Hemaniw in eneessl amount that WHO tu Falcosa moles of why - (0,084 - 0.99.62) moles Xcess moles moles 0.0018 molo Now concentration of NH3 in Gomb + 18 m2) = 79 mb Sobition = 0.0018 X1000. 0.023 M . ,11011 79..0.023 Mo "i calculation pH of weak basel Wriz » NH3 + H₂O NH3 + Ho lt VV021 WHA + SH Kb = [whairon 10 so, ko of Wriz = - kin 0.023-1 N ( let u amon l of whiz reach camount of, ons NI Lacts: 118x10 s = at - 00023-1 21 = x +18x10*)-01091x18"=0 x = 6.31 104 So, forl = 6.31 X10-9

so, por = -log [on] = -log (6.31x101) = 3.2 so, pH = 14- pOH = 14- 3-2 =