Molar mass of KHP ( potassium hydrogen pthalate) = 204.22 g/mol
TRAIL 1
mass of KHP taken = 0.4106 g
moles of KHP = mass /molar mass
= 0.4106 g / 204.22 g/mol
= 0.00020
NaOH(aq) + KHC8H4O4(aq) = KNaC8H4O4(aq) + H2O(l)
1 equivalent of NaOH = 1 equivalent of KHC8H4O4
moles of NaOH required to neutralize KHP = 0.00020
volume of NaOH = 19 55 ml
molarity of NaOH = moles x 1000/ volume in ml
= 0.00020 x 1000 / 19 55 ml
= 0.1028 M
HCl - NaOH titration
HCl + NaOH ----------------> NaCl + H2O
1 equivalent of HCl = 1 equivalent of NaOH
volume of NaOH = 21.80 ml
moles of NaOH = Molarity x volume in ml / 1000
= 0.1028M x 21.80 ml / 1000
= 0.00224 moles
1 equivalent of HCl = 1 equivalent of NaOH
volume of HCl = 25 ml
Molarity of HCl = moles x 1000 /25 ml
= 0.00224 moles x 1000/25
= 0.0896M
OR
Molarity of HCl = molarity of NaOH x volume of NaOH / volume of HCL
= 0.1028 M x 21.80 ml/ 25 ml
= 0.0896 M
TRAIL 2
mass of KHP taken = 0.4026 g
moles of KHP = mass /molar mass
= 0.4026 g / 204.22 g/mol
= 0.00197
NaOH(aq) + KHC8H4O4(aq) = KNaC8H4O4(aq) + H2O(l)
1 equivalent of NaOH = 1 equivalent of KHC8H4O4
moles of NaOH required to neutralize KHP = 0.00197
volume of NaOH = 19 20 ml
molarity of NaOH = moles x 1000/ volume in ml
= 0.00197 x 1000 / 19 20 ml
= 0.1027 M
HCl - NaOH titration
HCl + NaOH ----------------> NaCl + H2O
1 equivalent of HCl = 1 equivalent of NaOH
volume of NaOH = 22.15 ml
moles of NaOH = Molarity x volume in ml / 1000
= 0.1027M x 22.15 ml / 1000
= 0.00227 moles
1 equivalent of HCl = 1 equivalent of NaOH
volume of HCl = 25 ml
Molarity of HCl = moles x 1000 /25 ml
= 0.00227 moles x 1000/25 ml
= 0.0909 M
TRAIL 3
mass of KHP taken = 0.4065 g
moles of KHP = mass /molar mass
= 0.4065 g / 204.22 g/mol
= 0.00199
NaOH(aq) + KHC8H4O4(aq) = KNaC8H4O4(aq) + H2O(l)
1 equivalent of NaOH = 1 equivalent of KHC8H4O4
moles of NaOH required to neutralize KHP = 0.00199
volume of NaOH = 19 50 ml
molarity of NaOH = moles x 1000/ volume in ml
= 0.00199 x 1000 / 19 50 ml
= 0.1021 M
Average molarity of NaOH = (0.1028 M + 0.1027 M + 0.1021 M )/3
= 0.1025 M
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standard deviation =
Data the experiment Record For the lab report, attach this data sheet (signed by your TA) to your...
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