b. How many individual hydroxide ions (OH) are found in 13.4 mlL of0.586 M Ba(OH
ID: 580838 • Letter: B
Question
b. How many individual hydroxide ions (OH) are found in 13.4 mlL of0.586 M Ba(OH) (ag)? c. What volume (in L)of 0.586 M Ba(O1H) (ag) contains 0.466 ounces of Ba(OH), dissolved in i? d. If 16.0 mL of water are added to 31.5 mL of 0.586 M Ba(OHH): (a), what is the new solution molarity? e. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxides Ba(OH); (a) + 2 HC,Ho, (ag) Ba(C,H,0h (aq) + 2 Hon to a 5.00 mL sample of vinegar to reach the What volume (in ml.) of 0.586 M Ba(OH) (ag) must be added equivalence point? Use your average vinegar molarity (see page 1) in this calculation.Explanation / Answer
b) molarity of Ba(OH)2 = 0.586M = 0.586mol/L
Volume of Ba(OH)2 solution = 13.4ml
No of mole Ba(OH)2 = (0.586mol/1000ml)×13.4ml = 0.007852
No of mole of OH- ions = 0.007852× 2 = 0.015705
1mol = 6.022×10^23 ions/atom/molecule
No of individula OH- ions = 0.015705 ×6.022×10^23= 9.45×10^21ions
c) 1ounce = 28.35g
0.466 ounce = 0.466 × 28.35g = 13.211g
Molarity of Ba(OH)2 = 0.586 mol /L
Molar mass of Ba(OH)2(anhydrous) = 171.34g/mol
Mass of Ba(OH)2 in 1000ml = 0.586mol × 171.34g/mol= 100.41g
Volume of Ba(OH)2 solution contain 13.211g of Ba(OH)2 = (1000ml/100.41g)×13.211g =131.6ml
d) Total volume = 16ml + 31.5ml = 47.5ml
Dilution factor = 47.5ml/31.5ml=1.508
New molarity = 0.586M/1.508= 0.3886M
e) molarity of vinegar is not indicated
Calculate the volume as follows
i) Determine no of mol of Vinegar in 5ml
No of mol = (molarity/1000)×5
¡¡) Determine No of mole of Ba(OH)2 required
Mole of Ba(OH)2 required = No of mol of Vinegar /2
iii) Determine Volume of 0.586M Ba(OH)2 required
Volume of Ba(OH)2 should be added= (1000ml/0.586mol)×no of mol of Ba(OH)2 calculated
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