Question

Questions: 1. 495 mg of dried KHP is dissolved in 35 mL of distilled water and titrated with potassium hydroxide (KOH). If it took 22.02 mL of KOH to reach the endpoint, determine the concentration of KOH. Show calculations. If you forgot to dry the above KHP sample and it was later determined to contain 5.00 % water by weight, what would be the actual concentration of the KOH, taking into account the water content? Show calculations. 2. If you titrate 12.50 mL HCl with 23.37 mL of a 0.103 M NaOH to reach the endpoint, determine the concentration of the HCI. Show calculations. 3. If you standardized a NaOH solution against KHP, but, your buret was not clean as several drops of NaOH solution remained clinging to the inside of the buret, what effect would this have on your calculated concentration? Be specific, 4. After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.207 M. You perform five titrations and obtain the following results: 0.211, 0.204, 0.208, 0.199, and 0.197 M. a) What is the mean? b) What is the standard deviation? Exp 4 Analysis of an Acid by Titration

Answer 1

Q0 495 mg 1 aj mass of KAP 1039 Img = mass of KHP 0.495 g Molas mass of khP = R04.22_gmoêl given mass -0.495 g I no. of males of kupe Molar mass 204.22gma no. of moles of KHP = 2.42 x 103 mol trian AS Enckup КИСе Ицоч The fifcatioy reaction between kup and kon Кисе Ноу-+ Код 7 K₂ Ca Hydy +420 (1) (9) lae) (99) I mol of kHp Beact with 1 mol of KOH 2.42X16mal of 2:42X163 mol af komt kup requireel Thus we have No. of males of kHP = 2.42 xlos med Volume of KHP 29.02 mL

no al males [Conc] X1000 Volume-cmu) 3 & Conc. of Kot 2.42 Xco 82.2 9) Concenfation af koH = 0.109 m X1000 = 0+10.9-m 9m/ Ans Ans. b 6 gf KHP contain 5% water by weight So, it means 95% is dry KHP. So mass of küp= 0.495x95 loo Molar mats of kup= 204.729 mael no. of moles of KHP Given mass 0.470 8 malar mass 204.22 molt 10.470g - no. of moles of KHP 2.30 X163 mebel The tination reaction) Киселоу, tkon - K₂ Caty Oy there) (22) 199) (99) 1 mol of KHP required - 1 mol of kon

= 2.30X10² mol ef kHP 9.30-X173 mal af required KGH So no. of males af kon 2.30X163 mol Volume of KoH = 92.02mL * Conc. conc. of Koh no. of males af kon Xovo Volume cm) = conc. af koH 9.30 X103 x Too x २२.०२ Conce of Koh O. loym foi factual conc. of Kokl= 0.loy m Axs.