Many chemistry classes stop at analyzing pH change and what it means. However, there is another layer that requires calculus: related rates pertaining to pH change. The concepts of related rates taught in calculus classes can be expanded to pH as a solution evaporates, is diluted, or is titrated. This article explores problem-solving in this specific context. Note that the rate of change in pH is the slope of a pH/titration curve, the derivative of pH, which we will denote \(\frac{dpH}{dt}\).
###Generic Calculations###
Example 1: Prove that the pH and pOH of any solution change at rates with the same magnitude but in the opposite direction.
Solution: Mathematically speaking, we are referring to rates here. We want to prove that \(\frac{dpH}{dt} = -\frac{dpOH}{dt}\) for any solution. It's actually quite simple once you recall that, for any solution,
$$pH + pOH = 14$$
We can differentiate this equation. As \(pH\) and \(pOH\) are both functions of time:
$$\frac{dpH}{dt} + \frac{dpOH}{dt} = 0 \Rightarrow$$
$$\frac{dpH}{dt} = -\frac{dpOH}{dt}$$
This is generally a good statement to know, but if you undertsand the chemistry behind pH change, this is probably obvious. The introduction of derivatives, however, puts the idea into a new light for many people.
Now we can find a general statement for the rate of change of the pH in terms of the rate of change of the hydrogen ion concentration.
Example 2: By definition,
$$pH = -\log_{10}[H^{+}]$$
We can use a fancy differentiation technique. To keep the algebra clean, move the negative sign to the other side of the equation:
$$-pH = \log_{10}[H^{+}] \Rightarrow$$
Now take the base \(10\) exponent of both sides of the equation:
$$10^{-pH} = [H^{+}]$$
Now differentiate--differentiating exponents is easier than differentiating logarithms besides the natural logarithm. The left side requires the Chain Rule. Recall that \(\frac{d}{dx}a^x = \ln(a) \cdot a^x\), and combine this with the Chain Rule on the left-hand side:
$$\ln(10) \cdot 10^{-pH} \cdot -\frac{dpH}{dt} = \frac{d[H^{+}]}{dt}$$
Now isolate the rate of change of the pH:
$$\frac{dpH}{dt} = -\frac{10^{pH}}{\ln(10)} \cdot \frac{d[H^{+}]}{dt}$$
If you wish, you can use the same method to prove that
$$\frac{dpOH}{dt} = -\frac{10^{pOH}}{\ln(10)} \cdot \frac{d[OH^{-}]}{dt}$$
but the steps are essentially the same.
###General: Changes in Volume###
Now, recall that the concentration of a substance (in molars) is
$$\frac{molar \; amount}{Volume \; of \; solution}$$
Thus if either of these quantities changes, then the molarity changes. This introduces a new set of complications in the realm of related rates as you may know the rate of change of volume and must use that to find the rate of change of the molarity. As we said before:
$$[K] = \frac{m}{V}$$
where \([K]\) is the concentration of the solute, \(m\) is the mole quantity of the solute, and \(V\) is the volume of the solution. If the solution either evaporates or is diluted, then the volume changes, but the mole quantity is constant, so \(\frac{dm}{dt} = 0\). To find the rate of change of the concentration as the volume changes, we can use the quotient rule on the above equation to differentiate it. Again, the Chain Rule must be used on the functions of time (concentration and volume):
$$\frac{d[K]}{dt} = \frac{V\frac{d}{dt}(m) - m\frac{d}{dt}(V)}{V^2} \Rightarrow$$
$$\frac{d[K]}{dt} = \frac{V(0) - m\frac{dV}{dt}}{V^2} \Rightarrow$$
$$\frac{d[K]}{dt} = -\frac{m}{V^2}\frac{dV}{dt}$$
This expression can be substituted into the equation in the previous expression for a solution where the pH is measured:
$$\frac{dpH}{dt} = \frac{-m \cdot 10^{pH}}{\ln(10) \cdot V^2} \cdot \frac{dV}{dt}$$
This formula is used frequently in this subtopic but sufficiently infrequently that you may be better off remembering how to rederive this formula than you are trying to memorize it.
###Evaporations###
The first major category of pH changes are evaporations. In an evaporation, the volume of the solution decreases, but the mole amount of the solute remains the same, so its concentration increases. In an acidic solution, evaporation decreases the pH. However, in a basic solution, evaporation increases the pH. The related rates in these situations become even more complicated if you don't initially know the rate of pH change, but only the rate of volume change.
Example 3: A 50-liter solution of 0.5-molar hydrochloric acid is evaporating at a rate of .0005 Liters per second. At what rate is the pH changing when half of the solution has evaporated?
Solution: We need to know how many moles of hydrochloric acid exist in the solution:
$$50 \; L \; HCl \times \frac{0.5 \; mol \; HCl}{1 \; L} = 25 \; mol \; HCl$$
We can use this formula to find the rate of change of concentration:
$$\frac{d[HCl]}{dt} = -\frac{m}{V^2}\frac{dV}{dt}$$
We just found \(m\), and the other quantities on the right-hand side of the equation were given to us, so
$$\frac{d[HCl]}{dt} = -\frac{25 \; mol \; HCl}{(25 \; L)^2} \cdot (.0005 \; \frac{L}{s}) = .00002 \; \frac{M}{s}$$
Now we can use the fact that all of the hydrogen ions in the solution not formed in the equilibrium of water self-ionization are from the hydrochloric acid, so we let \([H^{+}] = [HCl]\). This substitution is always done in aqueous solutions--the ions formed from the water always revert back to water molecules--that's how the equilibrium works. We will use this step from now on without explicitly mentioning it.
Afterward, we use the equation for the rate of change of the pH.
$$\frac{dpH}{dt} = \frac{10^{pH}}{\ln(10)} \cdot \frac{d[H^{+}]}{dt}$$
Now, we need the pH of the solution at the exact moment in time we are analyzing. The initial concentration of hydrogen ions is \(0.50 \; M\), but the volume of the solution is halved due to evaporation, so the concentration doubles to \(1.00 \; M\). Then we find the pH:
$$pH = -\log_{10}(1) = 0$$
So,
$$\frac{dpH}{dt} = \frac{10^0}{\ln(10)} \cdot .00002 = 8.69 \times 10^{-6} \; \frac{pH}{s}$$
Example 4: A solution of sodium hydroxide has 22.5 moles of sodium hydroxide and has a starting pH of \(11.5\). The pH decreases at a constant rate of \(.025\) units per second as the solution evaporates. At what rate is the solution evaporating in liters per second when the pH of the solution is \(10.5\)?
Solution: In this case we are given the molar quantity of sodium hydroxide, the starting pH, and the rate of change of the pH. Thus we must work in the reverse direction relative to Example 3. We know that
$$\frac{dpH}{dt} = \frac{-m \cdot 10^{pH}}{\ln(10) \cdot V^2} \cdot \frac{dV}{dt}$$
We must find the volume of the solution when the pH is \(10.5\). That, however, requires finding the molarity of the solution, which in turn requires finding the \(pOH\) (as the solution is basic). The \(pOH\) is
$$pOH = 14 - pH = 14 - 10.5 = 3.5$$
Then we can find the molarity of the solution:
$$pOH = -\log_{10}[OH^{-}] \Rightarrow$$
$$3.5 = -\log_{10}[OH^{-}] \Rightarrow$$
$$[OH^{-}] = 3.16 \times 10^{-4}$$
This is important for finding the volume due to the definition of molarity:
$$[OH^{-}] = \frac{m}{V} \Rightarrow V = \frac{m}{[OH^{-}]} = \frac{22.5 \; mol \; OH^{-}}{3.16 \times 10^{-4} \; M} = 7.12 \times 10^{4} \; L$$
Now we can solve the rate of pH change equation for \(\frac{dV}{dt}\) and prepare to plug in all the variables. Just divide both sides of the equation by \(\frac{-m \cdot 10^{pH}}{\ln(10) \cdot V^2}\):
$$\frac{dpH}{dt} \cdot \frac{\ln(10) \cdot V^2}{-m \cdot 10^{pH}} = \frac{dV}{dt}$$
Finally, substitute everything:
$$\frac{dV}{dt} = .025 \cdot \frac{\ln(10) \cdot (7.12 \times 10^4)^2}{-22.5 \cdot 10^{10.5}} = -.004 \; \frac{L}{s}$$
That means the solution is evaporating at a rate of \(.004 \; \frac{L}{s}\). Do not worry about the negative sign; it is only present because the volume is decreasing.
###Dilutions###
A *dilution* is where the concentration of a solute is decreased by adding more solvent--typically water. Since the volume of the solution increases, the molarity decreases. Thus, for an acidic solution, the pH will increase, whereas for a basic solution, the pH will decrease. Thus in these problems, \(\frac{dV}{dt}\) will always be positive. The sign of \(\frac{dpH}{dt}\) depends on whether the solution is acidic or basic.
Example 5: A solution of barium hydroxide has a starting pH of \(12.00\) and contains \(.005000\) moles of \(Ba(OH)_2(aq)\). Water is added to the solution at a rate of \(1.5 \times 10^{-9}\) liters per second. When the pH of the solution has decreased to \(11.50\), at what rate is the pH decreasing?
Solution: Our target variable in this case is \(\frac{dpH}{dt}\).
First note that there are two hydroxide ions in every formula unit for barium hydroxide, so \(m = .010000\) moles, not \(.005000\) moles. We then center the calculations around the equation
$$\frac{dpH}{dt} = \frac{-m \cdot 10^{pH}}{\ln(10) \cdot V^2} \cdot \frac{dV}{dt}$$
The only other variable we don't have is the volume of the solution when the pH is \(11.50\), but that is easy enough to find, once we know the hydroxide ion concentration:
$$pOH = 14 - pH = 2.50 \Rightarrow$$
$$[OH^{-}] = 10^{-2.50} = .00316 \; M$$
We can solve for the volume:
$$.00316 \; M = \frac{.010000 \; mol}{V} \Rightarrow 3.16 \; L$$
With this value, we can substitute everything into the pH rate change equation and find \(\frac{dpH}{dt}\):
$$\frac{dpH}{dt} = \frac{-.010000 \cdot 10^{11.50}}{\ln(10) \cdot (3.16 \times 10^4)^2} \cdot 1.5 \cdot 10^{-9} = -.0021$$
This means that when the pH of the solution is 11.50, it is decreasing at a rate of \(-.0021\) units per second.
###Titrations###
The titration rate of change problems are the most complicated of the bunch, because you have the volume and the number of hydrogen/hydroxide ions changing. For this section we let \(\frac{dm}{dt}\) equal the number of moles of either hydrogen or hydroxide ions in the solution being titrated.
This problem analyzes a strong acid-strong base titration.
Example 6: 1200.0 milliliters of aqueous 0.10 molar hydrochloric acid are being titrated by 600.0 milliliters of aqueous 0.20 molar sodium hydroxide. When half of the basic solution has been titrated, at what rate is the pH of the solution changing? The base is being titrated at a rate of 10.0 milliliters per second.
Solution: First note that the solution will be getting increasingly basic, so \(\frac{dpH}{dt}\) will be positive. We also need to find the number of moles of hydrogen ions that will be present in the solution (apart from those formed in equilibrium with the water itself). To do this, we must find and understand the balanced chemical equation:
$$HCl(aq) + NaOH(aq) \rightarrow H_2O(l) + NaCl(s)$$
The reactants are used up in a \(1:1\) ratio. To find the number of moles of hydrogen ions in the solution (which is the same as the number of hydrochloric acid molecules) at the given point in time, we must find the amount of each reactant that is involved: the acid in the initial solution and the base in the part of the solution titrated up to this point.
Acid involved:
$$1200.0 \; mL \; HCl \times \frac{1 \; L \; HCl}{1000 \; mL \; HCl} \times \frac{0.10 \; mol \; HCl}{1 \; L \; HCl} = 0.12 \; mol \; HCl$$
Base involved:
$$300.0 \; mL \; NaOH \times \frac{1 \; L \; NaOH}{1000 \; mL \; NaOH} \times \frac{0.20 \; mol \; NaOH}{1 \; L \; NaOH} = 0.060 \; mol \; NaOH$$
Thus there are \(.060\) moles of excess acid, and now we can find the pH:
$$pH = -\log_{10}[H^{+}] = -\log_{10}\left(\frac{.060}{1.8000}\right) = 1.48$$
Now we can find the rate at which the hydrogen ion concentration is increasing:
$$\frac{d[H^{+}]}{dt} = \frac{m}{V^2}\frac{dV}{dt}$$
We know all of the values on the right-hand side:
$$\frac{d[H^{+}]}{dt} = \frac{.060}{1.5000^2} \cdot .01$$
Note that we converted the titration rate from milliliters per second to liters per second to maintain consistent volume units. This turns out as
$$\frac{d[H^{+}]}{dt} = 2.7 \times 10^{-4} \; \frac{M}{s}$$
With this information we can find the rate of change of the pH with the equation
$$\frac{dpH}{dt} = \frac{10^{pH}}{\ln(10)} \cdot \frac{d[H^{+}]}{dt}$$
Substitute the pH and rate of change of the hydrogen ion concentration.
$$\frac{dpH}{dt} = \frac{10^{1.48}}{\ln(10)} \cdot (2.7\ times 10^{-4}) = .0035$$