Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jeffmoonchop on March 27, 2024, 08:51:37 AM
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Hi all, I have a 20mM Tris pH 7.5 buffer and I am adding NaOH to adjust to pH 8.
In 1L, I am adding 2.1ml of 2N NaOH to reach pH 8. To calculate the concentration of NaCl produced, is it just as simple as:
Dilution factor 476.2
2000/476.2= 4.2mM NaCl?
thanks!
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I know that this is not an answer to your question, but why not just add Tris base to adjust the pH? EDT: Your calculation looks correct.
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I have 1M Tris pH 7.5 stocks and have been adding NaOH to adjust pH so I want to know how much NaCl I'm producing so I can compare this method to adjusting pH with Tris Base and its effect on lipid nanoparticle stability.
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4.2mM NaCl
Technically yes, that's the correct answer.
But I would just calculate number of moles of NaCl produced (equal to number of moles of NaOH) and divide it by the volume. Which is actually as trivial as possible, because the volume is 1 L. No need for combinations with dilution factor and whatnot.