I am doing analytical chemistry lab of preparing standard ac
I am doing analytical chemistry lab of preparing standard acid and base (titration)
I need the sample calculation of Naoh and HCL molarity each for their own 3 trials
Data for Naoh (I have to prepare Naoh by my self I believe molarity is 0.1)
trial KHP Naoh volume Naoh molarity
1 0.51g 33.82 mL ?
2 0.51g 30.64 mL ?
3 0.51g 31.12 mL ?
Data for HCL (I have to prepare HCL by my self I made a mistake by only use 500ml water instead 1000. I believe molarity is 0.2)
trial Na2CO3 HCL volume HCL molarity
1 0.13g 13.1 ?
2 0.13g 11.59 ?
3 0.13 9.87 ?
Please show calculations for all the molarity. Big thanks
Solution
trial KHP NaOH volume NaOH molarity
1 0.51g 33.82 mL 0.075 M
2 0.51g 30.64 mL 0.082 M
3 0.51g 31.12 mL 0.0803 M
Molarity = no. of moles of solute /volume in liters
Molarity = no. of moles of solute /volume in mili liters (1000)
no. of moles of solute = given mass / molar mass
KHP (Potassium hydrogen phthalate) molar mass = 204.2 g/mol
Trial 1 ; no. of moles of solute = 0.51 / 204.2 = 0.0025 M
Molarity = no. of moles of solute /volume in mili liters (1000)
Molarity = 0.0025 /33.82 ml (1000) = 7.4 (10-5)(1000) = 0.000075 (1000) = 0.075 M
Trial 2 ;Molarity = 0.0025 /30.64 ml (1000) = 8.2 (10-5) (1000) = 0.000082 (1000) = 0.082 M
Trial 3 ;Molarity = 0.0025 /31.12 ml (1000) = 8.03 (10-5) (1000) = 0.0000803 (1000) = 0.0803 M
Data for HCL (I have to prepare HCL by my self I made a mistake by only use 500ml water instead 1000. I believe molarity is 0.2)
trial Na2CO3 HCl volume HCl molarity
1 0.13g 13.1 0.092 M
2 0.13g 11.59 0.0103 M
3 0.13g 9.87 0.121 M
molar mass of Na2CO3= 106 g/mol
Trial 1 ; no. of moles of solute = 0.13 / 106 = 0.0012 M
Molarity = no. of moles of solute /volume in mili liters (1000)
Molarity = 0.0012 /13.10 ml (1000) = 9.2 (10-5)(1000) = 0.000092 (1000) = 0.092 M
Trial 2 ;Molarity = 0.0012 /11.59 ml (1000) = 1.03 (10-5) (1000) = 0.0000103 (1000) = 0.0103 M
Trial 3 ;Molarity = 0.0012 /9.87 ml (1000) = 1.21 (10-4) (1000) = 0.000121 (1000) = 0.121 M

