CHEM - VOLUMETRIC ANALYSIS

Scarbo

2011-07-31 12:53 am

1) 2.44g of barium chloridecrystals (BaCl2‧2H2O)were dissolved in water. To the solution was added 100cm3 of 0.125Mpotassium carbonate (K2CO3) solution. The precipitate formedwas filtrated, washed and dried. Calculate (a) the mass of barium carbonate (BaCO3) formed ANS:1.97g(b) the volumeof 0.10M nitric acid required to react completely with the excesspotassium carbonate in the filtrate. ANS: 50cm32) 1.00 g of calcium carbonate wasdissolved in 40.0cm3 of a hydrochloric acid solution. The excessacid required 25.00cm3 of a certain sodium hydroxide solution forneutralization.20.0cm3 of this sodium hydroxide solution neutralized16.8cm3 of the original acid. Calculate the molarities thehydrochloric acid and sodium hydroxide solution. ANS: 1.05M HCl , 0.880mNaOH
3) Y is a solution of a solid base XOH. 10cm3 of Y are added to 1000cm3of a 0.1M solution of sulphuric acid and the mixture requires 12.5cm3 of a 0.4Msodium hydroxide solution for complete neutralizion. What volume of Y isrequired to make 1 dm3 of 0.1 M solution of XOH? ANS:66.7 cm3

HOW CAN I FIND THE ANSWER? PLEASE GIVESOLUTION.THANKS!!

回答 (1)

pingshek

2011-07-31 8:48 am

✔ 最佳答案

1)
a)
BaCl2 + K2CO3 → BaCO3 + 2KCl
No. of moles of BaCl2•2H2O = 2.44/(137 + 35.5x2 + 1x4 +16x2) = 0.01 mol
No. of moles of K2CO3 = 0.125 x (100/1000) = 0.0125 mol (inexcess)
No. of moles of BaCO3 formed = 0.01 mol
Mass of BaCO3 formed = 0.01 x (137 + 12 + 16x3) = 1.97 g

b)
K2CO3 + 2HNO3 → 2KNO3 + H2O + CO2
No. of moles of excess K2CO3 = 0.0125 - 0.01 = 0.0025 mol
No. of moles of HNO3 = 0.0025 x 2 = 0.005 mol
Volume of HNO3 = (0.005/0.1) x 1000 = 50 cm³

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2)
CaCO3 + 2HCl → CaCl2 + H2O+ CO2
Volume of excess HCl = 16.8 x (25/20) = 21 cm³
Volume of HCl reacted = 40 - 21 = 19 cm³
No. of moles of CaCO3 = 1/(40+ 12 + 16x3) = 0.01 mol
No. of moles of HCl = 0.01 x 2 = 0.02 mol
Molarity of HCl = 0.02/(19/1000) = 1.05M

HCl + NaOH → NaCl + H2O
No. of moles of HCl = 1.05 x (16.8/1000) = 0.0176 mol
No. of moles of NaOH = 0.0176 mol
Molarity of NaOH = 0.0176/(20/1000) = 0.880M

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3)
"1000 cm³ of 0.1 M solution ..." should be "100 cm³ of 0.1 Msolution ..."

H2SO4 + 2NaOH → Na2SO4 + H2O
No. of moles of NaOH = 0.4 x (12.5/1000) = 0.005 mol
No. of moles of H2SO4 reacted with NaOH = 0.005 x (1/2) =0.0025 mol

H2SO4 + 2XOH → X2SO4 + H2O
No. of moles of H2SO4 reacted with XOH = 0.1 x (100/1000)- 0.0025 = 0.0075 mol
No. of moles of XOH = 0.0075 x 2 = 0.015 mol
Molarity of XOH in solution Y = 0.015/(10/1000) = 1.5 M

In 1 dm³ of 0.1 M XOH, no. of moles of XOH = 0.1 x 1 = 0.1 mol
Volume of solution Y needed = (0.1/1.5) x 1000 = 66.7 cm³

參考: micatkie

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