Question

Aav ν Α Α' 12 Ро Ever dy) V v V vab x, x' ADA Font Parac Student Handout for Titrations of Acids Laboratory CHE 121 Name: Date: A. Standardization of Sodium Hydroxide (NaOH) Solution Mass of bottle + KHP Mass of bottle Mass of KHP used Final buret reading Initial buret reading mL of NaOH used Molarity of NaOH Trial 1 1.3760 g 1.0369 g 0.2291 g 19.30 ml 4.90 mL 14.40 mL Trial 2 1.4045 g 1.0369 g 0.3676 g 35.60 ml 20.10 ml 15.50 ml Trial 3 1.3967 g 1.0369 g 0.3598 g 45.10 ml 30.10 mL 15.00 mL Average Molarity: Standard Deviation: Show your calculations for molarity and standard deviation. Moles KHP = The calculation of a molarity of NaOH (M) = The calculation average molarity for NaOH = The calculation of standard deviation (s) = B. Analysis of an Unknown Acid Trial 1 Mass of bottle + unknown 1.9762 g Mass of bottle 1.0317 g Mass of unknown used 0.9445 g Final buret reading 21.40 ml Trial 2 1.9557 g 1.0317 g 0.9240 g Trial 3 1.9813 g 1.0317 g 0.9446 g 2990 ml

Answer 1

TRIALI TRIAL2 TRIAL 0.2291g 0.3676g 0.35987 Mass of KAP useg volume of Naoh 14.40 ml 15.50ml. 15.oome NaOH a No moles of of NO moles of kHP. of NaOH t kHP Nakp +H₂O TRIAL 1 0.2291 g k HP 2 4 HP X 1 mol kHP Х 204, 2 g kHp. I mol Naolt ī mol UHP 0.001121 mol 2. Molarity of Nach 2 0.0719 M 2. 0.06112 0.0144L 2. TRIAL 2 No of Moles of NaOH. 0.36 76 g h HP, 204. 2g kthip 2. Orool8 Mol.

Molarity of Nadha 0.0018 mol 0.0155 L Woolt Orll6 M. 2 TRIAL - 3 moles No of of КНР Naolt 2 0.3598g KHP 204.2 g 2.0.00176 moles. 0.00176 moles 20.117 M O a 015 L Naat z Molarity Avarage Molarity B) Mi +Mz+Mz 2 3. O.lft 0.116 t 0.0779. 2 # 3. zoo 1037 M. 어 Caleulation of standard deviations *M, - Auarage Motarity M2 - Average moturity 20.1037 -0.116 2 0.1037 -0.0479. 20.0258. 2 -0.0123 0.1037 0.117. My A.M 2 20.0133 son Sof square of t standard devia

Standard deviation (0.0258)+(-0.01239++ (-0.0133) 3 0 0015 to.000176 0.00066 t z vo.0003. 2. 2 0.01732 got B) from the standardisation of Naott, we Molarity of Nadlh 20.1037'm 0.1037 moles of Nadt per 1 solution litre Means mmol Hence TATAL NaoH 2 of 0, 1037 mond Nach 1 ML soln TRIAL - volume mol of NaoH 320.35 ml. of waoH 220 XO 20.35 Xa10 37 m mol 1 ml. 2. 2.11 mmol. Immol KHO X g of KAP 2. 2011 mmol Nadh x 1 mmol Naolt Oo 2042 g RHP 0.43g Tamoch kHP Z.

69X. 89.027 1. HP X100 Avarage 0.43 0.483 %k HP 28KH x 100 8 sample TRIAL-2 NaOH 2 mmol NaoH volume of 2 19.85 ml mol of 2 19.85 X 0.1037m mo z. 2.05 mmol (Avarage (Avaraget 8 kHP 2 2.05 mmol of Naoh X 0.2042 g KHp (Avaraget z 0.4203g standar % RHP 2 0.4203 x 100 0.4717 289.10% TRIAL=3 volume of Nadh 2.20.40 ml. 2. 20.40 X 0.4037 mmol. 2.20115 mmol. mmol of Nadh g KHP. KHP 2.2.115 mmol of Nadth x 0.2042 2. On431883 g g % kHp 2 2 6, 431883 .9 X 100 0.4845 g z 89.14%

% KHP 89.02 + 89.10t 89014 2 Avarage 3. 89,086 4x10- z (Avarage to kHp - 7. kHp), ² tr 2560x10 3 3.6 X 10-3 (Average 1. KIHp - 4. K HP) (Avarage 1. KHP -% KHA 4. KHr) standard deviation z. X2 + 4x109 3.6 X103 2 3.6810 3 2. O,002 20.0447