172 ACIDITY 



Since the pH is the negative logarithm ^ of the molar H+ concentration, 



pH = - log (1.8 X 10-5) 



= - log 1.8 4- (-log 10-5) 



= - 0.26 + 5 



= 4.74 (Answer to Problem 1) 



Note that in this problem [H+] = K^, or pH = pKa. This relation 

 holds for any buffer where the acid and salt are present in equal amounts. 

 When different amounts are present, the pH may be calculated from 

 the equation: 



In the case of a base-type buffer, the corresponding equation is: 



[base] 



[OH-]=E:bX 



[salt] 



If the composition of the buffer and the numerical value of Kb are known, 

 [0H-] and, hence, pOH can be calculated. From this result the cor- 

 responding pH value can easily be found from the relation: 



pH + pOH = 14 



which holds for any aqueous solution at room temperature. 



The use of the above equations in buffer calculations is illustrated 

 below. 



Problem 2. What is the pH of 40 ml. of O.lM acetate buffer to which has 

 been added 10 ml. of O.liV HCl? 



The HCl added amounts to 1 m.e.q. (10 X 0.1), and the buffer originally con- 

 tained 4 m.e.q. (40 X 0.1) each of acetic acid and sodium acetate. For purposes 

 of calculation it may be assumed that the 1 m.e.q. of HCl reacts with 1 m.e.q. 

 of sodium acetate to form 1 m.e.q. of additional acetic acid": 



HCl + CHsCOONa > NaCl + CH3COOH 



Consequently, the acid : salt ratio of the buffer has been changed from 1 {i.e., 

 4 : 4) to 5 : 3. The pH may therefore be calculated by substituting known values 

 in the equation above: 



[H+] = 1.8 X 10-5 x-1 



= 3X 10-5 

 whence pH = 4.52 (Answer to Problem 2) 



1 The small letter "p" is used to mean "the negative logarithm of." 



2 More precisely, since the HCl, NaCl, and CHgCOONa do not exist as such in 

 water solution but are 100 per cent ionized at all times, the only reaction which 

 actually occurs is : 



H+ + CH3COO- -» CH3COOH 



