20 cm3 roztworu NaOH o pH=12 zmiareczkowano za pomocą 12,2 cm3 roztworu HCl. Obliczyć stężenie molowe kwasu.
NaOH ---> Na(+) + OH(-)
[NaOH]=[OH-]
pH + pOH = 14
pOH = 14-pH
pOH=14-12
pOH=2
pOH=-log[OH-]
[OH-]=10^-pOH
[OH-]=10^-2
[OH-] = 0,01mol/dm3
NaOH + HCl ---> NaCl + H2O
Cz*Vz/Ck*Vk = nz/nk
0,01*20/ Ck*12,2 = 1/1
0,2 = Ck*12,2
Ck = 0,0164mol/dm3
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NaOH ---> Na(+) + OH(-)
[NaOH]=[OH-]
pH + pOH = 14
pOH = 14-pH
pOH=14-12
pOH=2
pOH=-log[OH-]
[OH-]=10^-pOH
[OH-]=10^-2
[OH-] = 0,01mol/dm3
NaOH + HCl ---> NaCl + H2O
Cz*Vz/Ck*Vk = nz/nk
0,01*20/ Ck*12,2 = 1/1
0,2 = Ck*12,2
Ck = 0,0164mol/dm3