When two or more acids are present in a solution, the concentration of H+ (or pH) of the solution depends on the concentrations of the acids and their acidic constants Ka. The hydrogen ion is produced by the ionization of all acids, but the ionizations of the acids are governed by their equilibrium constants, Ka's.
Similarly, the concentration of OH- ions in a solution containing two or more weak bases depends on the concentrations and Kb values of the bases.
For simplicity, we consider two acids in this module, but the strategies used to discuss equilibria of two acids apply equally well to that of two bases.
When the contribution of pH due to self-ionization of water cannot be neglected, there are two equilibria to consider:
HA = H+ + A- C-x x x H2O = H+ + OH- 55.6 y y - - - ([H2O] = 55.6) Thus, [H+] = (x+y),Although you may use the method of successive approximation, but the formula to calculate the pH can be derived.
[A-] = x,
[OH-] = y,
and the two equilibrium constants are (x + y) x Ka = --------- - - (1) C - x and Kw = (x + y) y, - - - (2)
Solving for x from (2) gives
Kw x = ---- - y y and substituting this expression into in (1) results in (x+y) (Kw/y - y) Ka = ------------------- C - Kw/y + y Rearrange this equation to give: [H+] = (x+y) C - Kw/y + y = ------------- Ka Kw/y - y Note that Kw/y = [H+], and y = Kw/[H+]. Thus, we get: C - [H+] + Kw/[H+] [H+] = ---------------------- Ka [H+] - Kw/[H+] To simplify the formula, let h = [H+] Then, the formula becomes C - h + Kw/h h = --------------- Ka h - Kw/h
Now, consider the approximations we can make when we use this formula due to conditions.
C - h h = --------------- Ka h
h2 + Ka h - C Ka = 0 and -Ka + (Ka2 + 4 C Ka)(1/2) h = ----------------------- 2
C h = --------------- Ka h h = (C Ka)1/2
The treatment presented as Case 3 at the beginning of this tutorial is more general, and may be applied to problems involving two or more weak acids in one solution.
Assume x and y to be the concentrations of Ac- and Cc-, respectively, and write the concentrations below the equations:
HAc = H+ + Ac- 0.200-x x x HCc = H+ + Cc- 0.100-y y y [H+] = (x + y) Thus, you have (x + y) x ----------- = 1.8E-5 - - - (1) 0.200 - x (x + y) y ----------- = 1.4E-3 - - - (2) 0.100 - ySolving for x and y from (1) and (2) may seem difficult, but you can often make some assumptions to simplify the solution procedure. Since HAc is a weaker acid than is HCc, you expect x << y. Further, y << 0.100. Therefore, x + y => y and 0.100 - y => 0.100. Equation 2 becomes:
( y) y ----------- = 1.4E-3 - - - (2') 0.100 which leads to y = ((1.4E-3)(0.100))1/2 = 0.012 Substituting y in (1) results in (x + 0.012) x --------------- = 1.8E-5 - - - (1') 0.200 - xThis equation is easily solved, but you may further assume that 0.200 - x => 0.200, since x << 0.200. Thus,
-0.012 + (1.44E-4 + 1.44E-5)1/2 x = ----------------------------- 2 = 2.9E-4 - - - - Small indeed compared to 0.200You had a value of 0.012 for y by neglecting the value of x in (2). You can now recalculate the value for y by substituting values for x and y in (2).
(2.9E-4 + y) y ---------------- = 1.4E-3 - - - (2") 0.100 - 0.012 Solving for y in the above equation gives y = 0.011.You have improved the y value from 0.012 to 0.011. Substituting the new value for y in a successive approximation to recalculate the value for x improves its value from 2.9E-4 to a new value of 3.2E-4. Use your calculator to obtain these values.
Further refinement does not lead to any significant changes for x or y. Thus, your mission is accomplished.
You should write down these calculation on your note pad, since reading alone does not lead to thorough understanding.
For the solution of this problem, two methods are given here. If you like the x and y representation, you may use method (a).
Method (a) The two equilibrium equations are:
HA = H+ + A-; 0.0010-x x x H2O = H+ + OH- y y [H+] = (x+y) (x+y) x -------- = 4.0E-11 - - (3); 0.0010-x (x+y) y = 1e-14 - - - (4) Assume y << x, and x << 0.0010, then you have (x ) x -------- = 4.0E-11 - - - (3') 0.0010 x = ((0.0010)(4.0e-11))1/2 = 2.0E-7Substituting 2.0E-7 for x in 4 and solve the quadratic equation for y gives,
(2.0E-7+y) y = 1E--14 y = 4.1E-8Substituting 4.1E-8 in (3), but still approximate 0.0010-x by 0.0010
(x+4.1E-8) x -------------- = 4.0E-11 - - - (3'') 0.0010Solving this quadratic equation for a positive root results in
x = 1.8E-7 M <----- Recall x = [A-] [H+] = x + y = (1.8 + 0.41)1E-7 = 2.2E-7 pH = 6.65The next method uses the formula derived earlier.
Method (b)Using the formula from the exact treatment, and using 2E-7 for all the [H+] values on the right hand side, you obtain a new value of [H+] on the left hand side,
C - [H+] + Kw/[H+] [H+] = ---------------------- Ka [H+] - Kw/[H+] = 2.24E-7 pH = 6.65.The new [H+] enables you to recalculate [A-] from the formula:
(2.24E-7) [A-] = C Ka [A-] = (0.0010) (4.0E-11) / (2.24E-7) = 1.8E-7
Often, you may be attempted to use the approximation method:
x = (C Ka)1/2 = 2.0E-7 M A-, or H+; pH = 6.70and obtained a pH of 6.70.
The value 6.70 is greater than 6.65 by less than 1 %. However, when an approximation is made, you have no confidence in the calculated pH of 6.70.
However, the amount of H+ ions from water may be very small compared to the amount from an acid if the concentration of the acid is high.
When calculating [H+] in an acidic solution, approximation method or using the quadratic formula has been discussed in the modules on weak acids.
HAc = H+ + Ac- HCc = H+ + Cc- 0.200-x x x 0.100-y y y [H+] = (x + y)Which is larger, x or y?
Which is the stronger acid of the two? Most of the H+ will be from the stronger acid. Since HCc is a stronger acid (Kc =78 Ka), acidity is dominated by the ionization of HAc. Approximation to be made is to ignore x, i.e. x + y => y
You know that the pH is dominated by the stronger acid of the two. Based on that, the pH is 2.9. Can approximation be made?
At this concentration, the chloroacetic acid is completely ionized. The contribution from the self-ionization of water is still unimportant.