Question

2. Calculate the volume of potassium hydroxide at equivalence point and the pH at equivalence point when 10.00 16 of 1200 mM solution of acetic acid was titrated with 9.500 ml potass- sium hydoxide. pka value for acetic acid is 4.76. = 0.012 M acetic acid 12.00m M I M (millionlar) x 1000mm 9.500m IM * 1000 mm - 0.098M KOH mv = Mava of (0.012)(0.016) = 60.0095m) (V2) V2 = 0.012632 = 12.63 m2

Answer 1

Ansder Volume of 9.500 mM koH needed at equivalen ee point 12-63 mL (ca lcu lation is right . Total volume of solution at equivalenee paint (10.60+ 12.63) mL 22.63 mL Au of CHaca0H will be cn verted to cHa Cso at equivalen- -Ce point. So, PH at eguivalence Point depends on hydrolyAis of Ctaerao Moles of cH3coo Produced Moles of CH3CaH initially Present 12.00X10 10.00 moles 1.2X 10 moles Molarity of CHacaa at ewivalence point 1.2x 10-1 M (W) 01 XEOES 2.2.63 CH3o0 H20 CH3coDH + OH M): s.303x106 ci): +X +X -X E): 6.303x10)-x [eco0H LOHe [cH3 cra 2 (5.303x16)-x Ka(CHa croH) x2 1o 19 =S-75X1010 (5303x10)-x lo 4.76

x s.75X 1610)+ (515x1010)2+ (4x1X3.0 49x 1 x2) M *< (s303x 16)) 106 (axtbx+e-o> x = -bt - 4ac 2a 25.493 X1D M pH 14+l LoHO 14+x = 14+5-493 193XI0 8) 6.74 (species under thisd bracket repsesents R9 uilibriu m Cmeent rations)