Here is @Tobor the 8th Man original question
Can a soil slurry test show an alkaline ph due to high alkalinity water and the runoff ph be acidic at 5 to 5.5 at the same time?
If you water with water that has a total of 50-60 alkalinity does the soil just stay at that level? Same with 100 alkalinity? I am wondering if it builds up higher than what you put in?
If you had well water with ph 8.2 and ppm of 200 would total alkalinity be predictably high or could it still be in the 50-60 range?
Soils can accumulate elements that would lead to a higher outcome in a soil test, than is evident from the water supply or vice a versa. This oddity could be due to a loss of biology or humic content. There are lots of bacteria in water, there might be lots of fungus in soils, this might explain the variation, I would be fascinated to see this if it happening near you.
Here is @Tobor the 8th Man original question
Can a soil slurry test show an alkaline ph due to high alkalinity water and the runoff ph be acidic at 5 to 5.5 at the same time?
If you water with water that has a total of 50-60 alkalinity does the soil just stay at that level? Same with 100 alkalinity? I am wondering if it builds up higher than what you put in?
If you had well water with ph 8.2 and ppm of 200 would total alkalinity be predictably high or could it still be in the 50-60 range?
OK, so looking at this question, yes is the short answer, the whys and hows are largely found among the relationships of acids and the varying ranges of power, versus conjugate base, see the Henderson–Hasselbalch equation for more.
I have placed an example of the equation in working practice to help explain. I would still very much like to see this occurrence, since the shift is dramatic imo and worthy of investigation. It does sound counter intuitive, but its like a tug of war that has been unequally loaded on one side. The numbers are the same, but the outcomes will be different if the mass is not equal.
Hope this helps
Eco
If 10 mL of 1M NaOH are added to one liter of a buffer that is 0.3 M acetic acid and 0.2 M sodium acetate (Na+CH3COO–), how much does the pH change? The pKa for acetic acid is 4.76.
Answer:
In the original solution, acetic acid is the weak acid, and acetate is the conjugate base. Thus, [base] = 0.2 M, and [acid] = 0.3 M.
Using the Henderson–Hasselbalch equation,
pH = pKa + log [Base]
[Acid]
pH = 4.72 + log (0.2/0.3)
pH = 4.54
NaOH is a strong base, and dissociates completely. Therefore, adding the strong base results in
(0.01 liter)(1 M) = 0.01 moles of OH– ions into the solution.
All of the OH– ions will react with the acetic acid to form acetate ions:
CH3COOH + OH–
CH3COO– + H2O
Thus, when NaOH is added, the number of acetic acid molecules in the solution will decrease, but the number of acetate molecules in the solution will increase. In this case, 0.01 moles of acetic acid will be used up to neutralize the 0.01 moles of OH–, and this will form 0.01 moles of acetate ion in the solution.
At the end of the reaction:
Moles of acetic acid = moles of acetic acid before the reaction – moles of acetic acid used up in the reaction
= 0.3 moles – 0.01 moles = 0.29 moles
Moles of acetate = moles of acetate before the reaction + moles of acetate made in the reaction
= 0.2 moles + 0.01 moles = 0.21 moles
Now we are almost ready to plug back into the Henderson–Hasselbalch equation to find the new pH of the solution. We just need to remember that the volume has changed slightly, from 1 liter (1000 mL) to 1 liter plus 10 mL (1010 mL). The Henderson–Hasselbalch equation requires that the [Base] and [Acid] be molar (M, or mol/L) quantities. However, notice how that because both the acid and base species are in the same volume, the volume correction cancels out:
pH = pKa + log [Base]
[Acid]
pH = pKa + log (0.29 moles/1.01 liter)
0.21 moles/ 1.01 liter)
pH = 4.72 + log 1.38
pH = 4.86
The pH change is 4.86 – 4.54 = 0.32 units
For sure we can increase salts, this is not in question is it?