Question

A 27.8 mL sample of 0.292 M methylamine, CH₃NH₂, is titrated with 0.293 M hydroiodic acid.

At the equivalence point, the pH is .

Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

CH NH: molarity = 0.292M lk of CH NH = 4.4x104 volume = 27.8ml = 802782 1 (From google) Man pkb_leg kb_-10g (4.4x104) = 3.36 no of moles = 0.29200-0278 I pkw_14 -0.008 1176 at equivalence point moles of CH₃NH₂ = moles of HI = 0.0081176 molarity of HI = moles of HI volume of HI 0.0081176 0.293= volume of HI volume of HI = 0.027705L CH NHI CH NH₂ + HI -> weak base strong acid salt 0 I 0.0081176 0.0081176 -0.00 81176 C-0.0081176 70-0081176 6. 0081126 Total volume = o.0278 +0.027205= 0.055505 L y 0.0081176 LcH, H, 3 - -= 0.14625M = {] 0.055505 for this salt pH: formula: pH = £ { pk w _pkb_logc] = + {14 - 3.36-log (0.14625)] pH = 5.737

pH formula's for salt's weak accidt strong base ->salft ito pH = ± [pkw + pka + log c] strong acid & weak base salf tho pH = ₃ [ pk w plebe_loge