Question

If the K, of a monoprotic weak acid is 1.3 x 10 , what is the pH of a 0.27 M solution of this acid? pH =

Answer 1

Consider a mono protic weak acid 3 which dissociates in solution as: - HB - tt + Br The dissociation constant ka of this acid is ka [+][B ] ☺ thB where [++] = concentration of titions - [B ] = concentration of Brions [HB] = concentration acid HB

Dissociation constant is always expresse as a ratio of product of concentration of the products to the concentration of the reactants Now, let & be the concentration of ions which dissociate from HB then concentration of of u Brions would also be x as to dissociates into equal number H+ and 8 ions.

& [tb] = cox ; where c is the initial concentration of the acid. In this case, c = 0.27 So can be written as: - Ka= (x)(x) = taxt which gives - Kalc-x) = x² - КС – Кх - х? - О. or x² + Kax - Kal=0 (by taking negative a ...@ ign L con agu.commons The solution of this quadratic equation is given by x = - b ± √b²-4ac③ 2a On comparing with the standard equation axe²+bx+c=0 we a = b = kaca-kac So solution of becomes x= -kat (ka)²-4 24 (-Kal) (By (kalt (Bring

2 Thurs : - k +.C7) ____ Given Ka= 1:30 100 and c= 0.27m. se equation @ Desemes- - _ 2 - - 1. 3.X10 + N (I.3x10-6) + ५(1.2X16)(0.27) ___- - 1 - 3 x 106 + 1.69x101+ ). ५०५ x १०' 2- ___- - . .X10-6 2 .08 X 10-3 ____&o x- - 1. 3 x 10 " + 1. 18 x 10-30 - 1. 3 10-6- 1. 18 x 103 __ _ ( Neglecting the negative value of a as concentration cannot be negative so we get only one value of n) Thus X3 - 1:3x10-6 + 118x10-3 _* - 1. 18 x 10-3 ___ '2- -- 0.sax 10-3 The complex calculations can be done using scientific calculator)

Thus [++ ] = x= o.sax 10-3 M The formula to calculate pt of an acid is pt - log[it] So we get pH = -log(0.59 x103) $42-43.229) This pH = 3.229 - ~ 3:23