ACIDITY 



171 



The brackets indicate concentrations expressed on a molar basis. Since 

 weak acids ionize to only a slight degree, the numerical values of[H+] 

 and [A-] are small, whereas [HA] is large. Consequently K^ for weak 

 acids is a small number, for example, 0.000018 in the case of acetic 

 acid. The weaker the acid, the smaller is its Ka value, and vice versa. 

 Strong acids like hydrochloric are considered to be completely ionized 

 in water solution and, therefore, have no, or more exactly an infinitely 

 large, Ka value. 



Considerations exactly similar to those set forth above apply also to 

 weak bases. The corresponding expressions are: 



BOH^B+ + OH- 



and 



^ [B + ] • [0H-] 

 •^ [BOH] 



To illustrate how pH values of particular buffers may be calculated, 

 several typical problems will be worked out. 



Problem 1. What is the pH of a O.IM acetate buffer solution? 



The phrase "O.IM acetate bulfer" means that both acetic acid and sodium 

 acetate are present in O.IM concentration. The ionization equation for acetic 

 acid is: 



CH3COOH ?^ H+ + CH3COO- 



and its dissociation constant has the numerical value 1.8 X 10~^. The pH may 

 be found from the expression for the dissociation constant : 



[H + ] • [CH3COO-] 



A% = 1.8 X 10-s 



[CH3COOH] 



by inserting the proper values for [CH3COO-] and [CH3COOH], and solving 

 for [H+]. 



Let X=[H+]. Then [CH3COOH] = (0.1 - X), since the original acetic 

 acid concentration was O.IM. The concentration of acetate ions is the sum 

 of the concentrations resulting from the ionization of both acetic acid and sodium 

 acetate. That from the acetic acid is obviously X, while that from sodium 

 acetate is 0.1, because such salts are strong electrolytes and are completely ionized 

 in aqueous solution. Therefore, [CH3COO-] = U^l + A'). 



Substituting these values in the expression for K^we have: 



, X ' (0.1 + A^) 



1.8x10-== ^p,_^^' 



Now X is very much smaller than 0.1, since we are dealmg with a slightly 

 ionized acid, so (0.1 + A) and (0.1 - A") are both very nearly equal to 0.1. 

 Making this substitution, we have as a close approximation : 



, 0.1 A 

 1.8x10-5 = -^ 



or A = 1.8 X 10-5 



