Hussam Alsaoud ― still concerning to your post above: Worked example II ― On the pH of solutions containing a strong base (e.g. NaOH), besides, possibly, neutral salt(s) ― it is similar to worked example I (above): ― Determine the excess volume (Vexcess) of some unnamed strong base (MOH) aq. sol., of nominal molar concentration (formality to be precise), cMOH, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) basic pH. ― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pOH of the final basic solution would be: pOH = 14 - pH; pOH = -log10(f·cMOH/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pOH/{cMOH/(1 M)}. (*) Translates the base molar conservation.
Hussam Alsaoud ― still concerning to your post above: Worked example II ― On the pH of solutions containing a strong base (e.g. NaOH), besides, possibly, neutral salt(s) ― it is similar to worked example I (above): ― Determine the excess volume (Vexcess) of some unnamed strong base (MOH) aq. sol., of nominal molar concentration (formality to be precise), cMOH, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) basic pH. ― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pOH of the final basic solution would be: pOH = 14 - pH; pOH = -log10(f·cMOH/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pOH/{cMOH/(1 M)}. (*) Translates the base molar conservation.
I have a question, please. is the adding of NaCl at a specific concentration in buffer(more or less than the standard) effect on the properties like solubility, temperature or any kind of that??
I have a question, please. is the adding of NaCl at a specific concentration in buffer(more or less than the standard) effect on the properties like solubility, temperature or any kind of that??
Hussam Alsaoud ― concerning to your post above Worked example I ― On the pH of solutions containing a strong monoprotic acid (e.g. HCl), besides, possibly, neutral salt(s): ― Determine the excess volume (Vexcess) of some unnamed monoprotic strong acid (HA) aq. sol., of nominal molar concentration (formality to be precise), cHA, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) acidic pH. ― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pH of the final acidic solution would be: pH = -log10(f·cHA/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pH/{cHA/(1 M)}. (*) Translates the acid molar conservation.
Hussam Alsaoud ― concerning to your post above Worked example I ― On the pH of solutions containing a strong monoprotic acid (e.g. HCl), besides, possibly, neutral salt(s): ― Determine the excess volume (Vexcess) of some unnamed monoprotic strong acid (HA) aq. sol., of nominal molar concentration (formality to be precise), cHA, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) acidic pH. ― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pH of the final acidic solution would be: pH = -log10(f·cHA/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pH/{cHA/(1 M)}. (*) Translates the acid molar conservation.
In chemistry it is common to just use concentrations, but really we should use activities. Activity is effected by ionic strength and adding sodium chloride increases the ionic strength. This will lead to a pH change.
In chemistry it is common to just use concentrations, but really we should use activities. Activity is effected by ionic strength and adding sodium chloride increases the ionic strength. This will lead to a pH change.
The common ion effect of Na+ (introduced by NaCl) changes the equilibrium between monobasic and dibasic Na phosphate, causing a pH change.
The common ion effect of Na+ (introduced by NaCl) changes the equilibrium between monobasic and dibasic Na phosphate, causing a pH change.
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Hussam Alsaoud ― still concerning to your post above:
Worked example II ― On the pH of solutions containing a strong base (e.g. NaOH), besides, possibly, neutral salt(s) ― it is similar to worked example I (above):
― Determine the excess volume (Vexcess) of some unnamed strong base (MOH) aq. sol., of nominal molar concentration (formality to be precise), cMOH, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) basic pH.
― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pOH of the final basic solution would be: pOH = 14 - pH; pOH = -log10(f·cMOH/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pOH/{cMOH/(1 M)}.
(*) Translates the base molar conservation.
Hussam Alsaoud ― still concerning to your post above:
Worked example II ― On the pH of solutions containing a strong base (e.g. NaOH), besides, possibly, neutral salt(s) ― it is similar to worked example I (above):
― Determine the excess volume (Vexcess) of some unnamed strong base (MOH) aq. sol., of nominal molar concentration (formality to be precise), cMOH, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) basic pH.
― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pOH of the final basic solution would be: pOH = 14 - pH; pOH = -log10(f·cMOH/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pOH/{cMOH/(1 M)}.
(*) Translates the base molar conservation.
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Carlos Araújo Queiroz , Alastair Bain McDonald
thank you very much
Carlos Araújo Queiroz , Alastair Bain McDonald
thank you very much
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https://www.researchgate.net/post/The-effect-of-salt-concentration-on-the-pH-of-aqueous-solution
https://www.researchgate.net/post/The-effect-of-salt-concentration-on-the-pH-of-aqueous-solution
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I have a question, please.
is the adding of NaCl at a specific concentration in buffer(more or less than the standard) effect on the properties like solubility, temperature or any kind of that??
I have a question, please.
is the adding of NaCl at a specific concentration in buffer(more or less than the standard) effect on the properties like solubility, temperature or any kind of that??
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Yes, if chemical activity changes, it will make change in pH value.
Yes, if chemical activity changes, it will make change in pH value.
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I agree with N. Leyla Acan
I agree with N. Leyla Acan
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Hussam Alsaoud ,
I asked a similar question here: https://www.researchgate.net/post/How_much_acid_do_I_need_to_convert_a_basic_solution_into_an_acidic_soluion
The last answer by Henrik Rasmus Andersen is probably the clearest.
Hussam Alsaoud ,
I asked a similar question here: https://www.researchgate.net/post/How_much_acid_do_I_need_to_convert_a_basic_solution_into_an_acidic_soluion
The last answer by Henrik Rasmus Andersen is probably the clearest.
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Hussam Alsaoud ― concerning to your post above
Worked example I ― On the pH of solutions containing a strong monoprotic acid (e.g. HCl), besides, possibly, neutral salt(s):
― Determine the excess volume (Vexcess) of some unnamed monoprotic strong acid (HA) aq. sol., of nominal molar concentration (formality to be precise), cHA, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) acidic pH.
― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pH of the final acidic solution would be: pH = -log10(f·cHA/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pH/{cHA/(1 M)}.
(*) Translates the acid molar conservation.
Hussam Alsaoud ― concerning to your post above
Worked example I ― On the pH of solutions containing a strong monoprotic acid (e.g. HCl), besides, possibly, neutral salt(s):
― Determine the excess volume (Vexcess) of some unnamed monoprotic strong acid (HA) aq. sol., of nominal molar concentration (formality to be precise), cHA, that should be added to pure neutral salt(s) aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) acidic pH.
― Note that the said acidic solution would become diluted by the dilution factor f = Vexcess /Vfinal (*), where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pH of the final acidic solution would be: pH = -log10(f·cHA/(1 M)). Hence: Vexcess = Vfinal·f = Vfinal·10- pH/{cHA/(1 M)}.
(*) Translates the acid molar conservation.
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In chemistry it is common to just use concentrations, but really we should use activities. Activity is effected by ionic strength and adding sodium chloride increases the ionic strength. This will lead to a pH change.
In chemistry it is common to just use concentrations, but really we should use activities. Activity is effected by ionic strength and adding sodium chloride increases the ionic strength. This will lead to a pH change.
More
VOTE