Suppose you're washing out a bottle. You want to achieve the maximum dilution of the contaminants in the bottle for a given amount of water; a miser might phrase this as minimizing the amount of water needed to get the needed dilution of the contaminants.
If we abstract a little, we're repeating this process some number of times:
dumping all the water out of the bottle except for the little bit that won't come loose, which contains all the remaining contaminants; call that remaining water volume V0;
adding more water to dilute the water already in the bottle, up to some volume V1, diluting the contaminants by a factor V1/V0 by using an amount of water V1-V0, which we will then dump out when we repeat step 1.
After cycling some quantity V2 of water through this process, we've repeated it V2/(V1-V0) times, more or less, and achieved a dilution of (V1/V0)^(V2/(V1-V0)) times.
You might think that you don't want to go to extremes here. If V0 is one milliliter, and V1 is two milliliters, then on each cycle, you reduce the contamination by a factor of 3; so if you use twenty milliliters, you've reduced it by 3¹⁰, almost sixty thousand. By contrast, if you dump all twenty milliliters in at once, you achieve only a dilution of 1/21, which is pathetic; and if you go toward the other extreme and use one microliter of water on each rinse cycle, you go through twenty thousand rinse cycles, each providing a 1001/1000 dilution, which you might think wouldn't make much difference. But it turns out that that adds up to a dilution factor of 480 340 920 times. (You need the first thousand rinse cycles just to get to a dilution of roughly e, 2.717, but every thousand rinse cycles then adds up to another factor of e, and e²⁰ is almost five hundred million.)
That's about as good as you can do, though: rinsing with 20 times the amount of residual water that you can't get out will give you a dilution of up to e²⁰. And you can reach almost that optimum, three hundred million, say, with only 400 rinses, each of 50 microliters. At only 20 times the residual water, you don't get much additional benefit from reducing the amount of water on each rinse cycle below a twentieth or so of the residual water. However, because it's an exponential process, small differences in dilution will grow into larger ones as you repeat the process.
Which is to say that, perhaps unsurprisingly, a more or less continuous rinse is the ideal; you want to remove rinse water as fast as it's added. Allowing any amount of rinse water to accumulate that's significant compared to the residual, unremovable water will dramatically reduce the dilution you achieve. Once your effective number of rinses drops to 50, in the scenario above, you've lost more than an order of magnitude of dilution over 20 rinses.
Of course, if you have a way to reduce the residual water, that will be even more effective. Reducing the residual water is equivalent to increasing the rinse water by the same factor, which increases the exponent on e; in the above example, reducing the residual water by 50 microliters pumps up the optimum from e²⁰ to e²¹ (plus a bit), improving your final dilution by a factor of 2.7!
(This exponential sensitivity of the final result to the precise dilution ratio suggests use for sensitive measuring equipment. If you have a 1%-precision measure of the final concentration, say, you can derive an 0.05%-precision measure of the geometric-mean ratio of the ratio of the residual and rinse water in each step, or more if you repeat the dilution procedure more times.)
On the other hand, since it's an exponential process, the absolute magnitude of the problem is not so terrible.
The ideal is to achieve dilutions in the mole range (10²⁴), such that even if mole quantities of the contaminants were originally present, no molecules of them remain afterwards. (Succussion has not been shown to help.) The minimum to reach a dilution of 10²⁴, with the e²⁰ limit above, is when 20 ≈ 55.3; that is, you need 55.3 times as much rinse water as the residual water. If for some reason you're stuck in a low-efficiency regime where you can't avoid rinse-water accumulation, such that each rinse cycle only dilutes contaminants by a factor of 2, you need 80 rinse cycles, which means you need 80 times as much rinse water as residual water. At 55 rinse cycles, you will have only reached a dilution of about 3 × 10¹⁶, thirty million times more concentrated than with the optimal strategy.
That is, in more or less the worst case, your non-optimal rinsing strategy costs you only about 45% more water, even though when using the same amount of water, it's thirty million times worse.
(It's kind of amazing that it's feasible to achieve such extreme dilutions at all, let alone achieve them in your kitchen. Of course, there are always tricky details to get right, like that congealed grease spot full of PCBs that isn't getting diluted by your water. Don't get cocky.)
If your desire is not to remove every single molecule of whatever contaminants you have, but simply to reduce the pH of your laundry to the point where it won't deteriorate while drying or in storage due to the alkali, you have a much easier task. I'm going to guess that the pH of alkali laundry systems rarely goes above 10 and almost never goes above 11, and that reducing the pH to 8 before drying the laundry will provide adequate protection; that's a dilution of only 10³, which you can achieve with 10 rinse cycles in what I described as "more or less the worst case" and 7 "rinse cycles" in the best case of continuous mixing.
Welp, guess it's time to rinse this laundry.