Why does the pH increase after the alkali solution is diluted?
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Dear sea friends, I’m sorry – it was my mistake yesterday. I’m correcting it today: the alkaline solution mainly consists of Na2CO3 (3g/L) and NaHCO3 (30g/L), with a pH of around 8.5. When 2.5 ml of this solution is diluted in 50 ml of distilled water, the pH rises to around 8.9. Why does it increase? Seeking advice! This post was last edited by zjs0536 on 2009-3-13 12:09.]
Your alkali solution concentration is not high. I’m not sure what method you used to measure the pH value; I have two concerns: one is the measurement method, and the other is distilled water. It’s unlikely that distilled water has a high pH value; I think there is something wrong with the testing equipment or test strips. Just a suggestion, for your reference only!
It’s done with a pH meter, so there should be no problem; it’s the same instrument, the time difference isn’t significant, and it happens this way every time! Could it be that something else inside is causing the problem? Logically, even if there’s something else present, it shouldn’t lead to such a situation! I don’t understand!
This solution is weakly alkaline; Na2CO3 is a salt of a strong base and a weak acid. When diluted, the ionic equilibrium shifts in a direction that increases alkalinity, which is what causes the pH value to rise
Although the equilibrium shifts in the direction of increased alkalinity, the solution is also diluted, by about a factor of 20; overall, the pH should decrease. This explanation seems a bit forced!
Is there something wrong with your instrument? The alkaline solution has been diluted 20 times, which means the hydroxide ions have also been diluted 20 times; therefore, the pH value should decrease, right?
Then how can it be explained that the acidity of a concentrated acid is lower than that of a dilute acid? Within a certain range of dilution, OH- levels do not necessarily decrease. I need to take a look at chemical analysis when I have time; there are quite a few people interested in it here.
There can be no problem with the instrument; most likely, there is an issue with the solution, as some substance in it is causing the problem!
Have you measured the pH of the diluent? How much is it?
Poster: It seems the problem has been solved; in fact, just doing an experiment in the laboratory tomorrow will suffice. In the question you provided, the ratio of the two bases should be specified. This phenomenon should indeed exist – different compositions will yield different results when diluted
I’ll come and take a look at the specific figures tomorrow!
The stock solution is 8.5, and the diluted solution is 8.9
OP, are you from the Department of Chemistry and Environment at Hunan City University? Either the pH meter isn’t calibrated properly, or you made a mistake while recording the data. How can diluting an alkaline solution result in an increasing pH value? This post was last edited by lifanwang on 2009-3-13 09:39]
Is it because dilution increases the degree of hydrolysis of sodium carbonate? Either there were random errors during the measurement process
The pH of Na2CO3 and NaOH solutions is 8.5; I wonder why this concentration is so low? Based on what the poster has described, it seems to be a problem with the diluted solution
pH is related to the concentration of hydrogen ions; if the alkaline solution is too concentrated, its degree of ionization is low, and thus the pH value is low. After dilution, the base is fully ionized, resulting in a high pH value. I’m not sure if it’s right?
Now it can only be explained by enhanced hydrolysis, but it still seems a bit forced; I wonder if there is any theoretical data to support it!
It’s complicated to do the calculations; one needs to look up the respective ionization constants, and by using the concentrations of the solution before and after dilution, it’s possible to calculate the concentration of hydroxide ions. Once that is known, it becomes clear whether the cause is dilution or an instrument error.
The dissociation constants for carbonic acid are K1=4.3×10⁻⁷, pK1=6.37; K2=5.61×10⁻¹¹, pK2=10.25. Can someone help calculate these?