. What is the Chonical Bormula of the sebotonces present in each of the follawin
ID: 557567 • Letter: #
Question
. What is the Chonical Bormula of the sebotonces present in each of the follawing hosschold chonicals? (write onty the forswela) s Baking Soda b Epsom salts c Table sals d. Bleach e Houschold annonia A student weighed 0.76sg of mpure KIlC.lo and dissolved in 100 ml of distilled waer A fe drops of phenolphthalein indistor wese added and a solution of 0 100M sodium hydooxide was them S. titrated agaiast the KHCHOs untilled the indisator changed celor. The titration was repeated to obtain consistent titre values a showm in the tabble below. UIse the infornation and the table belaw to answer the questioess that follows Barette Rcading (m.) Final Burette Reading Initial Burete Reating 0.00 Tiere 36.03 6.00 Calculate the wverage titre a b. Write a balanced oquation Sot the reaction c Calculate the molarity of KHCsHO d. Determine the mass of KHCsHOs present in the sample (FW KC.H.Os-204.2 e. What is the percentage of KHCsO, present in the sample?Explanation / Answer
Ans. #8.a. Average titer = Sum of titer volumes / Number of trials
= (36.0 mL + 36.0 mL + 36.0 mL) / 3
= 36.0 mL
#b. Balanced Reaction: KHC8H4O4(aq) + NaOH(aq) -------> KNaC8H4O4(aq) + H2O(l)
#c. According to the stoichiometry of balanced reaction, 1 mol NaOH neutralizes 1 mol KHP.
So,
Moles of KHP in sample = Moles of NaOH consumed to reach endpoint
Or, Moles of KHP in sample = (Molarity x Average titer volume) of NaOH
Or, Moles of KHP in sample = 0.100 M x 0.036 L = 0.0036 mol
Hence, Moles of KHP in sample = 0.0036 mol
Now,
Molarity of KHP in initial solution = Moles of KHP / Initial volume in liters
= 0.0036 mol / 0.100 L
= 0.036 M
#d. Mass of KHP in sample = Moles x Molar mass
= 0.0036 mol x (204.2 g/ mol)
= 0.73512 g
#e. % KHP in sample = (Mass of KHP / Mass of sample) x 100
= (0.73512 g / 0.765 g) x 100
= 96.09 %
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