The rate at which water flows out of a biuret
is directly proportional to the amount of water, *W*, remaining above the
exit valve:

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ln W = kt + C, where C is a constant.

But at t = 0, W = 54.3 ml. This applies to a 50.0 ml biuret.
The volume is greater than 50.0 ml because there is an additional 4.3 ml
between the 50.0 ml line and the valve.

ln(54.3) = k(0) + C.

C = ln(54.3) = 3.994524227

If it takes 300 s for 50.0 ml of
water to flow out of the burette, then there will be only 4.3 mL remaining:

ln(4.3) = k(300) + 3.994524227

k = -0.008453030680

ln W = -0.008453030680
t + 3.994524227

In exponential form:

W = e^{-0.008453030680 t + 3.994524227}

It is more convenient to measure the volume, V, flowing out of the
burette. V = 54.3 – W, or substituting,

V = 54.3 - e^{-0.008453030680 t + 3.994524227}