Question

What is the percent ionization of a monoprotic weak acid solution that is 0.179 M? The acid-dissociation (or ionization) constant, K,, of this acid is 2.32 x 10-

Answer 1

HAT Ho - Hot + I o.129 Elo-129- T +X X 1-1 X 1 X [Hot] [A] Ka [HA] 2-32x104, X 0.179- 2-32 x10" (0-179-x) = x2 0.41528x10 - 2.32 x 10 (x) = x² x2 + 2.32x10(X) - 0-41528810ll zo ax² + bxtc.20 -b7b²-400 x= za - 2.32x2011 I / 62-32x1011) ²-4186-0.41528X1011) -2-32 xlot! [(2.32*2.32x17") + (2x0.41528) 11511 X + 0.00000 40757 -2.32 XLO x= 1 - 2.32 x1811 + 0.00000 40752 X2 - X2 0.0000020378 M ionized concentration (X) Xloo ionization - initial concentration t 0.0000 20378 XLOO 0.179 = 0.0011.38 %

HAT Ho - Hot + Á I 0.1791-0101 -x + x 0-124-X X X [Hz ot] LAI Ka I HA] x 2.32x loll 0.179-* 2-32 xlõ" (0-179-x)=x? 0.41528xloll -2.32 Xoll ex)=x² x²+2.32x10" (x) - 0-41528 xloll zo ax2 + bx+c 0 - bb49c x= za - 2032 xloll +(2-32617") 4(17 (-0.41528 X1011) - 2017 -2-32 xroll + ][12.32 2-32 X17") +(4X0-4152817161 xia 1-2-34 X10 + 0.00000 40257 - 2-32 XLOH + O-COCCO 40752 X> - 0.00000 20378 M ionized concentration (X) Xloo ionizaltma - initial concentration 0.000020378 X100 = 0.001138 40 0.179