Question

Molarity of KHP: _0.500 Volume of KHP: 25mL How many moles of KHP are in the flask? 1L 0.500 moles - 0.0125 moles KHP 25mL KHP- How many grams of KHP is in the flask? (MW - 204.22g/mol) 0.0125 moles XHP.med = 2.55g KHP Desired NaOH Concentration: Trial #1 2.75 0 Trial #2 2.75 0 Trial #3 2.75 0 33.05 33.01 32.97 Initial pH of KHP Initial Buret Reading Final Buret Reading Volume of NaOH Delivered Initial (H+] Concentration Molarity of NaOH Deviation of Molarity 33.05 33.01 32.97 2.67e-14 2.67e-14 2.67e-14 0.379 0.378 0.379

not sure if molarity of NaOH was calculated correctly and also unsure of how to find deviation

Answer 1

your molarityvalues are correct but your initial [H+] are not correct.

since initial pH = 2.75

and pH = - log [H+] ,

so [H+] = 10-pH = 10 -2.75  = 1.778 x 10-3 M

question :

number of moles of kup = 0.0125 moles kup since kup & Naon reacts in 1:1 ratio , so for Trial 1) concentration x 1 mol Naon x = 0.0125 mol KHP L volume 9 Naou Imol Kup (in L) of Naon und volume of Naои има = 33.05 m2 = 0.03305 L 0.3782 mol/L Holarily of Naon 0.0125 moles - 0.03305 L = 0.3782 M Trial 2 : = concentration (Molavily of Naon) 0.0125 moles 33.01 m2 0.0125 moles 0.03301 L = 0.37807 M Trial 3] Trial 3 0.0125 moles concentration (Holarity of Naon) 0.0125 moles 32.97 m2 0.03297L = 0.3791 M Average mol qily = 0.3782 + 0.3787 + 0.3791 M - = 0.37867 M * Deviation from molarilly = theoilical Desired molquilly Desired molarity = 0.400M experimental value of motarily Trial 1 Trial 2 Trial 3 Deviation (0.400 -0.3702) 6.400-0-3787) = 0.0213 M (0.4-0.3791) = 0.0209 M molarity = 0.0218 M 0.0218 + 0.0213 + 0.0209 Average deviation S of molauilty 3 : 0.0213 M % everear = (theoritical value - experimental value) X 100 theoritical value 10.4-0.37867) x 100 o.u = 5.33%

let us assume that we spill out 0.2 g of KHP while weighing.

mass of KHP = 2.55 - 0.2 = 2.35 g

number of moles of KHP now would be = 2.35 g / 204.22 g mol-1 = 0.011507 moles

molarity of NaOH for trail 1 would be = 0.011507 moles / 0.03305 L = 0.34818 M

so the experimentally determined molarity of NaOH would decrease.

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