Mark M. answered 10/20/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let P = pH = -0.4343(lna)
Given: dA/dt = 0.003 and a = 0.02
Find: dP/dt
dP/dt = -0.4343(1/a)(dA/dt) = -0.4343(50)(0.003) ≈ 0.065
The pH is increasing at the rate of approximately 0.065 units/minute when a = 0.02 and dA/dt = 0.003.