Answer to Question #73499 in Inorganic Chemistry for miksongoal

Question

Calculate the effect on the ph after the adding of 0.006mol of HCL at 300ml a buffer which is 0.250M in CH^(3)COOH and 0.560M in NAHCH^(3)COO.

Answer 1

Initial:

n(HCl)=0.06 mol

n(CH3COOH)=0.25M*0.3l=0.075 mol

n(HCH3COONa)=0.56M*0.3l=0.169 mol

pH=pKa+log(C(CH3COONa)/C(CH3COOH))=4.76+log(0.56/0.25)=5.11

Final:

CH3COONa+HCl=NaCl+CH3COOH

n(HCl)=0 mol

n(CH3COONa)=0.169mol-0.06mol=0.109 mol

n(CH3COOH)=0.075 mol+0.06mol=0.135 mol

C(CH3COOH)=n/V=0.135/0,3=0.45M

C(CH3COONa)=n/V=0.109/0,3=0.363M

pH=pKa+log(C(CH3COONa)/C(CH3COOH))=4.76+log(0.363/0.45)=4.67

Answer: pH change from 5.11 to 4.76