Standardization of Sodium hydroxide (NaOH) solution with Potassium hydrogen phthalate (KHC2H404)
Table B: Titration for knowing concentration of unknown HCI solution:
(Titration of HCl vs. NaOH)
(Titration of KHCSH:04 vs. NaOH)
S. No
Trial 1
SN
1.
2.
3.
4.
Experiment Data
Molarity of KHP (nol)
Initial burette reading of NaOH (MI)
Fmal burette reading of NaOH (ml)
Final Volume of NaOH dispensed (m) (3-1)
Trial
01M
Q0 ml
9.6 TL
9.0
Trial 2
QUIM
0.0 mL
9.8 ml
9.5 mL
Volume of rid HCl solution (m)
10.00 mL
Intial burente reading of NaOH (m2
0.0 ml
Final burette reading of NaOH (ml)
103 m.
Volume of NaOH dispenyed (3-2)
10.3 ml
Mola concentribon of NaOH (mol). From table A, using 02104 M
Fortula balow
Trial 2
10 0 TL
0.0 ml
9.7 ml
9.1 ml
0.104 M
Calculations: of Trial 1
CALCULATION:
Chemical Reaction:
Equi molar reaction
1 H2O (1)
+ 1 NaCl
1 KHC8H4O3(aq) + 2NaOH(aq)
Chemical Reaction1NaOH (OCT + 1 HCI (aq)
(Salt)
1 H2O(l) +1 NK H404
Acid
Base
Water
Salt
Use the same Formula:
Use the same formula : MIV1]KHP = M2V2]NaOH
MI = molarity of HCI=?
M1 molarity of KHP=0.1 M
Vi volume of HCl =
V1 volume of KHP =
M2 =molarity of NaOH =
M2 molarity of NaOH = ?
V2 volume of NaOH(burette reading) =
V2 Volume of NaOH (burette reading) =
MIV1]acid = [MV2]b.
M1 V1 = M2 V2
M1(HCI) = M2V2/V1
Calculate:
Screens 1-2 of 2