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Reaction Rate Kinematics and pH Buffering (Chemical Engineering) Solution

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Problem Overview

We evaluate the time sensitivity of an industrial acidity neutralization process (d(pH)dt\frac{d(\text{pH})}{dt}) governed by a general logarithm sensor function and a product-exponential concentration profile:

pH(H+)=−log⁡10(H+),H+(t)=t2e−0.5t\text{pH}(H^+) = -\log_{10}(H^+), \qquad H^+(t) = t^2 e^{-0.5t}

Step-by-Step Mathematical Solution

Step 1: Deconstruct via the Chain Rule

Because pH is a function of hydronium concentration H+H^+, and H+H^+ is a function of time tt:

d(pH)dt=d(pH)dH+⋅dH+dt\frac{d(\text{pH})}{dt} = \frac{d(\text{pH})}{dH^+} \cdot \frac{dH^+}{dt}

Step 2: Differentiate the Outer Logarithmic Layer

Recall that ddx[log⁡10(x)]=1xln⁡(10)\frac{d}{dx}[\log_{10}(x)] = \frac{1}{x\ln(10)}. Therefore:

d(pH)dH+=−1H+ln⁡(10)\frac{d(\text{pH})}{dH^+} = -\frac{1}{H^+ \ln(10)}

Step 3: Differentiate the Inner Concentration Layer via Product Rule

The function H+(t)=t2e−0.5tH^+(t) = t^2 e^{-0.5t} is a product of u=t2u = t^2 and v=e−0.5tv = e^{-0.5t}:

  • u′=2tu' = 2t
  • v′=−0.5e−0.5tv' = -0.5e^{-0.5t} (by the Exponential Chain Rule)

Applying the Product Rule:

dH+dt=(2t)e−0.5t+t2(−0.5e−0.5t)=e−0.5t(2t−0.5t2)\frac{dH^+}{dt} = (2t)e^{-0.5t} + t^2(-0.5e^{-0.5t}) = e^{-0.5t}(2t - 0.5t^2)

Step 4: Combine Layers via Chain Rule and Simplify

d(pH)dt=(−1H+ln⁡(10))⋅(e−0.5t(2t−0.5t2))\frac{d(\text{pH})}{dt} = \left(-\frac{1}{H^+ \ln(10)}\right) \cdot \left(e^{-0.5t}(2t - 0.5t^2)\right)
Substitute H+(t)=t2e−0.5tH^+(t) = t^2 e^{-0.5t}:
d(pH)dt=−e−0.5t(2t−0.5t2)(t2e−0.5t)ln⁡(10)=−2t−0.5t2t2ln⁡(10)=0.5t2−2tt2ln⁡(10)\frac{d(\text{pH})}{dt} = -\frac{e^{-0.5t}(2t - 0.5t^2)}{(t^2 e^{-0.5t})\ln(10)} = -\frac{2t - 0.5t^2}{t^2 \ln(10)} = \frac{0.5t^2 - 2t}{t^2 \ln(10)}
Factor out tt for t>0t > 0:
d(pH)dt=0.5t−2tln⁡(10)\frac{d(\text{pH})}{dt} = \frac{0.5t - 2}{t \ln(10)}

Step 5: Evaluate Operating Conditions at t=4 minutest = 4\text{ minutes}

d(pH)dt∣t=4=0.5(4)−24ln⁡(10)=2−24ln⁡(10)=04ln⁡(10)=0 pH units/min\frac{d(\text{pH})}{dt}\Big|_{t=4} = \frac{0.5(4) - 2}{4\ln(10)} = \frac{2 - 2}{4\ln(10)} = \frac{0}{4\ln(10)} = 0\text{ pH units/min}

Engineering Conclusion

At t=4 minutest = 4\text{ minutes}, the rate of change of pH is exactly zero (d(pH)dt=0\frac{d(\text{pH})}{dt} = 0). This identifies a critical stationary point where the chemical effluent reaches its maximum acidity (minimum pH) before exponential dilution reverses the trend. Process control engineers program automated PID buffer valves to detect this zero-crossing so that injection pumps safely transition from ramp-up mode to steady-state holding, avoiding chemical overdosing.