Question

Consider the titration of 100.0 mL of 0.100 M methylamine (CH3NH2) with 0.500 M HNO3. Calculate the pH at the following volumes of acid added.

For CH3NH3+, pKa = 10.632

(a) Find the equivalence point volume.

(b) 0 mL

(c) 9.0 mL

(d) 10.0 mL

(e) 20.0 mL

(f) 30.0 mL

Answer 1

Jhn we itate weaase CCHNH) with stronq acid The e, u a hon for titra hon İS s ea Given dah CH3NH2 concen ha hon (Cb)-Ο. (oort CH3NH> HNO3 Concen Por CH3NHj pka - 10.632- Accor eu'ng to sto ichi»metry of neutraliza han at euivalence peint nacil- h 1 5 Now lels in pal eawivalence point at euvalence [cH3NH]- se 6 x o.loo isassocia LcHa NHa']- o.os3 M uilibrium acid is Pa

io ka-4-28 는 2.33x10 [cH3NHJ x 2 OX o o s 3-х = 2.33%10-11 1. 3X 10 ( 5. 851) (b)PH at omL acid mix Now it ls only bqse (Weak base) Cb : 0.100 Ka2.33 x11 Ka 2.3 3x 10 11 We icalculate POH and then will convert to pt .OH POH 2.69 So PH: 14 - PoH 14- 2.63- 11 1.31 Pa 9e 2

Untitita on tyui valence point is rehed the molu added of acid aie rea«ted with pase finding numbe, of mola f base * adde acid. We can knouo how nuch base is vemainini. h ew +otal volume we can f'nd ta hon.f b molel o Concen ase Ow ust uaカon [<H3 NH ^ we can fnd poH b which we can Addution ef 10.omL HNoj Atelence point 5 5 As n qeston aso ewvalence point its γeaahed so r" can be calculated like ueshon Cc). Adolu hon o oin\ aahivement. is e^valen* point ach H be samu as calculatel in jueshon Gj) Paqe 3