10. (EXTRA CREDIT) In a titration experiment, a student prepared a sample by dis

Question

10. (EXTRA CREDIT) In a titration experiment, a student prepared a sample by dissolving 0.2512 KHP (MW -204.22 g/mol) and titrated with a NaOH solution of unknown concentration. It was found that 12.05 mL of NaOH solution was used to reach the end point when phenolphthalein was used as the indicator. Next, this student used this calibrated NaOH solution to titrate an unknown acid. The unknown acid solution was prepared by dissolving 0.3214 g of unknown acid in approximately 50 m water, then a few drops of phenolphthalein was added. It was found that 24.53 mL of NaOH solution was used to reach the end point. Calculate the molecular weight of the unknown acid, given that both KHP and unknown acid reacts with NaOH in 1:1 molar ratio. The unknown acid used is one of the three: KHC4H 06, KHCgOs, KHC 04. Based on MW, identify the unknown acid. (10 points) 1127.30

Explanation / Answer

m = 0.2512 g KHP

mol of KHP = mass/MW = 0.2512 /204.22 = 0.00123 mol of KHP

KHP + NaOh raito is = 1:1

mol of KHP = mol of NaOH

mol of NaOH = 0.00123

[NaOH] = mol/V = 0.00123 / (12.05*10^-3) = 0.102074 M

then...

let the unknown aicd be HA

m = 0.3214 g of HA, V = 50 mL of HA

V require d= 24.53 mL

mol of NaOH = MV = 0.102074*(24.53*10^-3 ) = 0.0025 mol of acid

since ratio is

1:1

mol of HA = 0.0025

MM = mass/mol = 0.3214/0.0025

MM = 128.56 g/mol of aicd

nearest value is KHC2O4,

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