CHEM - VOLUMETRIC ANALYSIS

Scarbo

2011-07-31 1:07 am

1) A water soluble compound, MHCO3,reacts with hydrochloric acid according to the equationMHCO3(aq)+HCl(aq)à MCl(aq)+CO2(aq)+H2O(l)It was found that 25.0cm3 of a solution containing 10.8gdm-3 of MHCO3 reacted with 27.0cm3 of ahydrochloric acid solution of concentration 3.65g dm-3.Calculate a) the molarity of the hydrochloric acid ANS: 0.100M b) the molarity of the MHCO3 solution ANS:0.108Mc) the molarmass of MHCO3 ANS: 100g d) the atomicmass of M. ANS: 39 2) 1.26 g of a ydrateddibasicorganicc acid of anhydrous molecular mass 90were made up to 250cm3 withwater. 25.0cm3 of this solution required 20.0 cm3 of 0.100M potassium hydroxidesolution for neutration. Clculate the number of molecules of water ofcrystallization per molecule of the hydrated acid. ANS:2 3) Sodium hydroxide solution isadded in excess to 2.0g of a mixture of ammonium sulphate and sodium sulphate. Oheating, the ammonia liberated is dissolved in 25cm3 of 0.5Msulphuric acid. The excess acid required 27.0 cm3 of 0.1 M sodiumhydroxide for complete neutralization. Calculate the percentage composition ofthe mixture. ANS: 73.6%(NH4)2SO4,26.4% Na2SO4 4) 3.60g of a solid mixture of sodium carbonate-10 water(Na2Co3‧10H2O) and sodium hydrogencarbonate(NaHCo3)required 23.0cm3 of 1.50M hydrochloric acid for complete reaction. Calculatethe percentage by mass of each substance in the mixture. ANS: 47.3% Na2CO3‧10H2O, 52.7% NaHCo3

回答 (1)

不用客氣

2011-07-31 7:57 am

✔ 最佳答案

1)
a)
Molarity of HCl = 36.5/(1 + 35.5) = 0.100M

b)
MHCO3 + HCl → MCl + H2O+ CO2
No. of mol of HCl = 0.1 x (27/1000) = 0.0027 mol
No. of mol of MHCO3 = 0.0027 mol
Molarity of MHCO3 = 0.0027 / (25/1000) = 0.108 M

c)
Mass of MHCO3 = 10.8 x (25/1000) = 0.27 g
Molar mass of MHCO3 = 0.27/0.0027 = 100 g (or 100 gmol⁻¹)

d)
Atomic mass of M = 100 - (1 + 12 + 16x3) = 39

2)
Let H2A•nH2O be the acid.

H2A + 2NaOH → Na2A + 2H2O
No. of mol of NaOH = 0.1 x (20/1000) = 0.002 mol
No. of mol of H2A in titration = 0.002 x (1/2) = 0.001 mol

Total no. of mol of H2A•nH2O = 0.001 x (250/25) = 0.01mol
Molar mass of H2A•nH2O :
90 + n(1x2 + 16) = 1.26/0.01
n = 2

3)
H2SO4 + 2NaOH → Na2SO4+ H2O
No. of mol of NaOH = 0.1 x (27/1000) = 0.0027 mol
No. of mol of excess H2SO4 = 0.0027 x (1/2) = 0.00135 mol

2NH3 + H2SO4 → (NH4)2SO4
Total no. of mol of H2SO4 = 0.5 x 25/1000 = 0.0125 mol
No. of mol of H2SO4 reacted with NH3 = 0.0125- 0.00135 = 0.01115 mol
No. of mol of NH3 = 0.01115 x 2 = 0.0223 mol

(NH4)2SO4 + 2NaOH → Na2SO4+ 2H2O + 2NH3
No. of mol of NH3 formed = 0.0223 mol
No. of mol of (NH4)2SO4 = 0.0223 x (1/2) =0.01115
Mass of (NH4)2SO4 in the mixture = 0.01115 x (14x2+ 1x8 + 32+ 16x4) = 1.472 g
% by mass of (NH4)2SO4 in the mixture =(1.472/2) x 100% = 73.6%
% by mass of Na2SO4 mixture = 1 - 73.6% = 26.4%

4)
Let m g be the mass of Na2CO3•10H2O in themixture.

Na2CO3•10H2O + 2HCl →2NaCl + 11H2O + CO2
No. of mol of Na2CO3•10H2O = m/(23x2 + 12 +16x13 + 1x20) = m/286 mol
No. of mol of HCl used = (m/286) x 2 = m/143 g

NaHCO3 + HCl → NaCl + H2O+ CO2
No. of mol of NaHCO3 = (3.6 - m)/(23 + 1 + 12 + 16x3) = (3.6 - m)/84
No. of mol of HCl used = (3.6 - m)/84

Total number of moles of HCl used :
(m/143) + [(3.6 - m)/84] = 1.5 x (23/1000)
m = 1.701

% by mass of Na2CO3•10H2O in the mixture =(1.701/3.6) x 100% = 47.3%
% by mass of NaHCO3 in the mixture = 1 - 47.3% = 52.7%

參考: micatkie

👍 0 👎 0