# Message Boards

Answer
(Unmark)

GROUPS:

6

# Approximations for calculating pH & their qualitative meaning

Posted 3 months ago

Approximations for calculating pH & their qualitative meaning by Rauan kaldybaev
Introduction Out[]=
As a physicist learning chemistry, one thing that bothered me about how we calculate the pH of a solution is that we usually don’t account for the dissociation of water. It is true that in most practical applications, dissociation of water can be neglected. Still, conceptually it somehow seems wrong to not take water into account. If there are two chemical reactions going on in a solution, it is only proper to think of both of them and only neglect one of them when it’s completely clear that it can be neglected. In addition, the “standard formula” for acid in water problems predicts that the pH pH
Deriving the equation of chemical equilibrium of acid in water Consider a solution containing a small amount of acid. For definiteness, let’s restrict the discussion to monoprotic acids, which can only donate one + H HCl H 2 H 3 + O - A HCl H 2 H 3 + O - Cl [ H 3 + O - A [HA] K a ( 1 ) Because nothing enters or leaves the solution and the radical - A - A [HA]+[ - A a 0 a 0 [HA]+[ - A a 0 ( 2 ) Apart from the dissociation of acid, another process that goes on in the solution is the dissociation of water, 2 H 2 H 3 + O - OH [ H 3 + O - OH K w ( 3 ) The ion H 3 + O 2 H 2 H 3 + O - OH H 2 H 3 + O - A [ - OH d [ H 3 + O - A - OH ( 4 ) Combining the two previous equations, we arrive at the formula for the concentration of H 3 + O [ H 3 + O a+ 2 a K w 2 ( 5 ) Here, a≡[ - A pH≡-Lg[ H 3 + O pH=-Log10 a+ 2 a K w 2 ( 6 ) If no acid dissociates, a=0 H 3 + O K w -7 10 pH a K w [ H 3 + O H 3 + O a a 0 K w 2 K a 2 a 0 a K a a 0 ( 7 ) This equation follows directly from the chemical equilibrium equations for the dissociation of acid and water (1), (3) and from the molar equations (2), (4). So far, it does not involve any approximations. And although it could be solved exactly using Cardano’s formula, the latter is far too complicated to give much quantitative insights into the problem and too bulky for hand calculations. Luckily, the answer to (7) can be, to a good degree of precision, found using several simple approximations discussed in the next sections.
K w In most practical applications, the quantities a a 0 K a K w a a 0 K w 2 a a 0 2 K a 2 a a 0 K a a 0 a a 0 2 a a 0 ( 8 ) For K w 0= 2 a K a K a a 0 a≈ 2 K a a 0 K a K a 2 H 3 + O H 3 + O a≈ 2 K a a 0 K a K a 2 K w 2 K a a 0 K a K a 2 ( 9 )pH K a a 0 a≈ a 0 K a a 0 K a a≈ a 0 a[ a 0 a 0 a 0 pH a 0 pH [ H 3 + O pH≡-Log10[ H 3 + O a 0 a [ H 3 + O a [ H 3 + O [ H 3 + O pH≡-Log10[ H 3 + O Out[]=
We can see that as a 0 K w a a= a approx ε<<1 a approx K w a approx K w ε≈ -2 a 0 K a K a a 0 K a -4 2 a 0 K a a 0 2 K a K a K a a 0 K a K w K a K a a 0 K a 2 K a K w K w The formula can be significantly simplified if a 0 K a K w ε≈ K w 2 a 0 K a ( 10 ) Substituting this into the formula (6) for pH a 0 K a K w pH ΔpH≈ 1 2Log[10] K w a 0 K a ( 11 )Out[]=
As you can see, error decreases as 1/ a 0 H 3 + O H 3 + O pH 0.005 K a -5 10 pH 2.4· -8 10 pH a 0 K a -14 10 H 3 BO 3 K a -10 10 -4 10 pH pH≈ pH K w 1 2Log[10] K w a 0 K a ( 12 ) The error of pH K w 1/ a 0 1/ a 0 pH pH pH pH Out[]=
Supplemental code In[]:= Simplify- #/.ε0 D[#,ε]/.ε0 2 KA 2 (a0-a) 2 a 2 a 2 KA 2 Out[]= -2a0-KA+ KA(4a0+KA) Kw-4 2 a0 3 KA 5 KA 2 KA KA(4a0+KA) -4Kw-2KAKw+2KA(4a0+KA) KwIn[]:= Asymptotic -2a0-KA+ KA(4a0+KA) Kw-4 2 a0 3 KA 5 KA 2 KA KA(4a0+KA) -4Kw-2KAKw+2KA(4a0+KA) KwOut[]= Kw 2a0KA In[]:= LimitSimplifyDD-2a0-KA+ KA(4a0+KA) Kw-42 a0 3 KA 5 KA 2 KA KA(4a0+KA) -4Kw-2KAKw+2KA(4a0+KA) Kw,Kw/.a01 x Out[]= 1 2KA In[]:= Simplify a+ 2 a 2 2 KA 2 Kw 2a0KA Out[]= - Kw a0KALog[100] In[]:= Block Kw= -14 10 -9 10 #+ 2 # 2 2 a a0 2 KA 2 a a0 a a0 2 a a0 #+ 2 # 2 2 KA 2 |