a solid sample of impure Ba(OH02 is added to .4000L of .5000M of HBr. the remaining solution was acidic, it was then titrated to equivalence point with .1200L of .5000M NaOH. what mass of pure Ba(OH)2 was present in the sample?
The idea here is that an excess of HBr was added to Ba(OH)2. Some reacted with Ba(OH)2 and some was left over. How much was left over? That amount is what was titrated with the NaOH.
Ba(OH)2 + 2HBr ==> 2H2O + BaBr2 +(xs HBr)
mmoles HBr initially = 400.0 x 0.5000M = 200.0 mmoles.
Amount back titrated = 120.0 mL x 0.5000M NaOH = 60 mmoles.
Amount HBr used up with the Ba(OH)2 reaction is 200.0-60.0 = 140.0 mmoles.
Looking at the reaction, mmoles Ba(OH)2 = 1/2 mmoles HBr = 1/2*140.0 = 70 mmols Ba(OH)2 that reacted.
grams Ba(OH)2 = moles x molar mass = ??
the grams of Ba(OH)2 that reacted was 11.99g. but isn't this still of the impure Ba(OH)2 sample? how do i find out the mass of pure Ba(OH)2?
thanks for you're help so far
No. That's the mass of Ba(OH)2 in the sample, unless of course, the HBr titrated something else in addition to the Ba(OH)2. If we knew the mass of the impure sample, and we don't, the percent Ba(OH)2 could be determined.
%Ba(OH)2 in the sample = (11.99/mass sample)*100 = ??
To find the mass of pure Ba(OH)2 present in the sample, we can follow these steps:
Step 1: Determine the balanced chemical equation for the reaction between Ba(OH)2 and HBr.
Ba(OH)2 + 2HBr → BaBr2 + 2H2O
Step 2: Calculate the moles of HBr that reacted with the impure Ba(OH)2 using the given volume and concentration.
moles of HBr = volume of HBr solution (L) × concentration of HBr (M)
Given:
Volume of HBr solution (V1) = 0.4000 L
Concentration of HBr (C1) = 0.5000 M
moles of HBr = 0.4000 L × 0.5000 M
Step 3: Use stoichiometry to relate the moles of HBr to the moles of Ba(OH)2.
From the balanced equation: 1 mole of Ba(OH)2 reacts with 2 moles of HBr.
moles of Ba(OH)2 = (moles of HBr) / 2
Step 4: Calculate the moles of Ba(OH)2 that reacted with HBr.
moles of Ba(OH)2 reacted = (0.4000 L × 0.5000 M) / 2
Step 5: Determine the moles of NaOH used in the subsequent titration. Since NaOH and Ba(OH)2 react in a 1:1 ratio at the equivalence point, the moles of NaOH used will be the same as the moles of Ba(OH)2 that reacted.
moles of NaOH = concentration of NaOH (M) × volume of NaOH solution (L)
Given:
Volume of NaOH solution (V2) = 0.1200 L
Concentration of NaOH (C2) = 0.5000 M
moles of NaOH = 0.1200 L × 0.5000 M
Step 6: Calculate the mass of pure Ba(OH)2 using the moles of Ba(OH)2 reacted.
mass of Ba(OH)2 = moles of Ba(OH)2 reacted × molar mass of Ba(OH)2
The molar mass of Ba(OH)2 can be calculated as follows:
Ba: 137.33 g/mol
O: 16.00 g/mol (two oxygens)
H: 1.01 g/mol (four hydrogens)
molar mass of Ba(OH)2 = (1 × 137.33 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol)
By substituting the values in the equation, you can find the mass of pure Ba(OH)2.