Question

Calculate the pH at the equivalence point when 30 mL of 0.293 M (CH3)2NH, is titrated with 0.693 M HBr.

Answer 1

(CH3)2NH + HBr ---> (CH3)2NH·HBr

no fo moles of CH3CH2NH2 = 0.293M x 0.03 L = 0.00879 moles

from the balanced equation one equilent of (CH3)2NH required one equilent of HBr

0.00879 moles of CH3CH2NH2 required 0.00879 moles of (CH3)2NH

and 0.00879 moles of salt will form

volume of HBr = 0.00879 mol / 0.693 mol/L

= 0.0127 L

= 12.7 mL

total volume = 30 + 12.7 = 42.7 mL

concentration of (CH3)2NH·HBr = 0.00879 / 0.0427 = 0.15 M

pKa of (CH3)2NH = 10.77

now use the direct formula

pH = 1/2 (pKa -logC)

pH = 1/2 (10.77 - log0.15)

pH = 5.8