Question

Consider the titration of 40.0 mL of 0.0600 M (CH3)2NH (a weak base; Kb = 0.000540) with 0.100 M HCI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (b) 6.0 mL (c) 12.0 mL (d) 18.0 mL

Answer 1

B) after 6.0 ml

(Base = Salt ) which show below

$\frac{M_1×V_1}{V_1+V_2}= \frac{M_2×V_2}{V_1+V_2}$

$(Base)=\frac{0.06×40}{40+6}=0.0522M$

$(Acid)=\frac{0.1×6}{40+6}=0.013M$

Base left = $0.0522-0.013=0.0392 M$

$POH=Pkb+ log(\frac{salt}{base})$

$POH=3.36+ log(\frac{0.013}{0.0392})$

$PH=14-POH=14-2.88=11.12$