a solid sample of impure Ba(OH)2 is added to 400.0 mL of 0.5000 M of aqueous HBr, a strong acid. the remaining solution was acidic. it was then titrated to equivalence point with 120.0 mL of 0.5000 M of NaOH solution. what mass of pure Ba(OH)2 was present in the sample?
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To determine the mass of pure Ba(OH)2 present in the sample, we need to understand the reaction that occurs between Ba(OH)2 (the impure sample) and HBr (the strong acid). From the given information, we know that Ba(OH)2 reacts with HBr to form water (H2O) and BaBr2.
The balanced chemical equation for the reaction is as follows:
Ba(OH)2 + 2HBr → 2H2O + BaBr2
Now let's break down the problem step by step:
1. Calculate the amount in moles of HBr present initially:
Given concentration of HBr = 0.5000 M
Volume of HBr solution = 400.0 mL = 0.4000 L
Using the formula: moles = concentration × volume
moles of HBr = 0.5000 M × 0.4000 L = 0.2000 mol
2. Since the reaction between Ba(OH)2 and HBr is a 1:2 ratio, we can determine the number of moles of Ba(OH)2 that reacted:
From the balanced equation, we see that 1 mole of Ba(OH)2 reacts with 2 moles of HBr.
Therefore, moles of Ba(OH)2 = 0.2000 mol ÷ 2 = 0.1000 mol
3. Next, we need to find the concentration of NaOH:
Given concentration of NaOH = 0.5000 M
Volume of NaOH solution used = 120.0 mL = 0.1200 L
Using the formula: moles = concentration × volume
moles of NaOH = 0.5000 M × 0.1200 L = 0.0600 mol
4. Since NaOH reacts with BaBr2 in a 1:1 ratio, and BaBr2 is formed from the reaction between Ba(OH)2 and HBr, the moles of NaOH used in the titration are equivalent to the moles of Ba(OH)2 initially present. Therefore, moles of Ba(OH)2 = 0.0600 mol.
5. Calculate the molar mass of Ba(OH)2:
Ba = 137.33 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of Ba(OH)2 = (137.33 g/mol) + 2 × (16.00 g/mol + 1.01 g/mol) = 171.35 g/mol
6. Calculate the mass of pure Ba(OH)2 in the sample:
mass = moles × molar mass
mass = 0.0600 mol × 171.35 g/mol = 10.28 g
Therefore, the mass of pure Ba(OH)2 present in the sample is 10.28 grams.