Taylor R-0870 test powder (crystals) - hard to measure for anyone?

I got the Taylor K-2006 test kit this season (first time pool owner) and from day 1 the R-0870 powder has had a good bit of large-ish crystals in it. Is this typical? I feel like the large/chunky crystals make it difficult to accurately measure the amount when testing. Also, the powder seems to stick to the dropper which makes it doubly hard to measure as I know I'm not getting an accurate amount.

I could get something to grind the crystals if needed. Doesn't seem like I should need to do that but not a big deal. And do others of you have to regularly clean the dropper?

You don't need to be accurate. The instructions say to use two dippers but one works fine most of the time. One clue here is thst the instructions say to use two dippers for either the 10ml sample or the 25ml sample.

Just put some in, if the sample turns pink, you've got chlorine.

I just tap the dipper on the top of the test vessel and let the R-870 corner of my kit box get a little messy.

I noticed the same thing and also see the crystals seem to sit in the water sample no matter how much I swirl. Is that okay? I'm concerned it will affect my test somehow.

The refill container I got prior to the one I'm using now was the same way--it came to me already in crystals. You're right, a lot of it does stay undissolved in the water but that's okay. If you get pink, then don't worry about what doesn't dissolve. Your test will still be accurate.

So if I get any pink when I add the powder/crystals (regardless of qty of powder/crystals) I will get an accurate reading? That surprises me - but pleasantly surprised if that's the case.

I'm no scientist, but I had been wondering if the crystals all dissolved, would my sample have been a darker pink and would it have taken more drops to turn it clear and therefore would I get a larger FC reading?

That's what I was thinking. I would think the # of drops of the reagent in comparison the amount of the powder is important - not just in comparison to the presence of the powder. But I also am not a scientist.

Think of it this way:
The powder color tags the Free Chlorine. It doesn't matter how much because the Free Chlorine takes the color it needs, the color is just there for the ride.
The titrating drops eliminate the color tagged Free Chlorine. When there's no more color tagged Free Chlorine the color goes clear.
The 5 drops for the CC test turn any amount of Combined Chlorine into Free Chlorine which then picks up the color it needs.
Then you use the titrating drops again to eliminate the newly Freed Free Chlorine.

Does this make sense?
I'm probably wrong on the chemistry but I think the analogy works.

The amount of DPD is not critcal. You are doing a titration. It works like this:
1) DPD is added to the sample. If any free chlorine (or free bromine if you are testing bromine) is present it oxidizes the DPD and forms what is known as Wurster Dye which is a magenta colored molecule. If there is a really high level of chlorine prenent then this Wurster Dye is futher oxidize to a colorless imine (the bleachout that is seen with DPD) adding more DPD power will shift this chemical reation to reform the Wuster Base (only possible with FAS-DPD testing and not with DPD testing because with DPD testing we are looking at how deep the pink color is, i.e. the AMOUNT of Wurster Base formed in our sample. With FAS-DPD testingi we just need it as an indicator so we know when we have neutraized all the FC in the sample)
2)we then titrate with a reducing agent (Ferric Ammonium Sulfate--FAS) which 'neutalizes' the chorine in the sample. THIS is the quantitative part of the test. The amount of FAS used is important, not the amount of DPD. Once all the FC is neutralized and gone then there is non left to react with the DPD and the Wurster Dye is gone also--the sample turns colorless)
3) since this test is specific for hypochorous acid and hypochlorite ion (free chlorine or the equivalent acid and ion for bromine or iodine ) we can then determine the amount of other oxidizer present (usually our chloramines or bromamines but also this test will show MPS and ozone present) by adding a potassium iodide solution. This is easily oxidized by all the above oxidizing agents (chloramines, bromamines, MPS, ozone) and forms the iodine equivalents to hypochlorous acid and hypochlorite ions (Free Iodine) that once again form Wurster Dye and turn our sample pink again. If there is no oxidizer present the sample stays clear and we have no CC (or MPS or Ozone interference). If it turns pink we titrate again to reduce the free iodine back into iodide ions and the sample turns from pink to colorless and we have our CC reading.

So you see, the amount of DPD is not critical. You just need enough to determine if there is FC (or Free Bromine or Free Iodine) present in the sample and then you convert that into either chloride, bromide, or iodide by titrating witth a reducing agent (FAS). When all the FC (or other free halogen) is all converted there is no longer any left to react with the DPD and form Wurster Dye and the sample is colorless again. The amount of FAS needed to do this is what measures your FC (or Free Bromine)

To determine the CC you convert it into Free Iodine and titrate again. You do this by adding potassium iodide (which is very easily oxidzed into free iodine). The FAS reduces the free iodine back into iodide and, once again, the amount of FAS needed determines the concentration. The DPD is only a means to create the Wurster Dye, which is our indicator. The AMOUNT of Wurster dyd created is determined by the amount of Free halogen present and not by the amount of DPD, which is added in excess (even 1 scoop is excess as long as the pink color forms!)
Hope this explains it.

(Yeah, I'm the other chemistry geek in the forum!:cool:)

Someone on another forum did a test using different scoops and they did get slightly different readings but generally only differing by 1 drop or sometimes 2 if the transition was on the edge between drops. The reason is mostly that with less dye the transition from light pink to clear is a little more subtle. There is an equilibrium between the chlorine in the water and that which has reacted with the dye (though the reaction is mostly towards reacting with dye) so with more dye the intensity of color at any given chlorine level (even a low level) is higher, though not proportionately. So it is potentially more likely for someone to miss the transition because it is so faint (i.e. they think something is clear when it is really just very faint pink). In other words, the issue isn't with the chemistry, but with your eyes and how clearly you can see more subtle transitions in color.

In practice, I use just one heaping scoop in a 25 ml sample and don't have a problem seeing the transition, but that's already a 0.2 ppm resolution so not a big deal. For those using a 10 ml sample, the transition is much stronger anyway since one drop represents 0.5 ppm so again, not a big deal.