# Estadistik

Páginas: 2 (307 palabras) Publicado: 15 de octubre de 2012
CALCULOS
FORMULAS
PH=-log [H+] [H+]=((V NaOH)(C NaOH)-(VHCL)(C HCL))/((V HCL+V NaOH))[HCl]=((V HCL)(C HCL)-(V NaOH)(C NaOh))/((V HCL+NaOH))

[HCl]=((20)(0.0939)-(5)(0.10429))/((20+5))=0.054
PH= 1.267[HCl]=((20)(0.0939)-(10)(0.10429))/((20+10))=0.0278
PH=1.555

[HCl]=((20)(0.0939)-(14)(0.10429))/((20+14))=0.0122
PH= 1.913

[HCl]=((20)(0.0939)-(16)(0.10429))/((20+16))=0.005815PH=2.235

[HCl]=((20)(0.0939)-(17.5)(0.10429))/((20+17.5))=0.005815
PH=2.815

[HCl]=((20)(0.0939)-(17.6)(0.10429))/((20+17.6))=0.001130PH=2.946

[HCl]=((20)(0.0939)-(17.7)(0.10429))/((20+17.7) )=0.0008505
PH=3.070

[HCl]=((20)(0.0939)-(17.8)(0.10429))/((20+17.8) )=0.00029653PH=3.527

[HCl]=((20)(0.0939)-(17.9)(0.10429))/((20+17.9) )=0.00029575
PH=3.52

[HCl]=((20)(0.0939)-(18)(0.10429))/((20+18) )=0.000020526
PH=4.68

[HCl]=((18.5)(0.10429)-(20)(0.0939))/((20+18.5) )=1.880
PH=-0.274

[HCl]=((18.10)(0.10429)-(20)(0.0939))/((20+18.10) )=1.838
PH=-0.264[HCl]=((18.15)(0.10429)-(20)(0.0939))/((20+18.15) )=20.380
PH=-1.309

[HCl]=((18.20)(0.10429)-(20)(0.0939))/((20+18.20) )= 1.848
PH=-0.266

[HCl]=((18.25)(0.10429)-(20)(0.0939))/((20+18.25) )=1.854
PH=-0.268

[HCl]=((18.30)(0.10429)-(20)(0.0939))/((20+18.30) )=1.815
PH=-0.258[HCl]=((18.35)(0.10429)-(20)(0.0939))/((20+18.35) )=1.865
PH=-0.270

[HCl]=((18.40)(0.10429)-(20)(0.0939))/((20+18.40) )=1.870
PH=-0.271

Regístrate para leer el documento completo.