An aqueous solution of HNO3 has a PH of 2.30 what will be the new PH if 10 mL. of this solution is diluted by 30 mL. of H2O ?
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Nice Chemistry Question
Started by heshyn2000
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heshyn2000 said:
pH = 2.9
PH= -log[H+]
2.3 = -log[H+]
[H+] = .005
N1V1= N2V2 (N=M Monoprotic)
.005 * 10 = N2 * 40
N2=M2= .00125
New PH= -log .00125
PH= 2.9
2.3 = -log[H+]
[H+] = .005
N1V1= N2V2 (N=M Monoprotic)
.005 * 10 = N2 * 40
N2=M2= .00125
New PH= -log .00125
PH= 2.9
dentalengineer said:
where is 0.005 from?
2.3 is the -log of .005 or 5X10 -3joonkimdds said:
dentalengineer said:
Thanks. Explanations are quite useful.
