Zn(OH)2 solid (molecular mass = 99.41 amu) is added to 0.500 L of 0.50 M HBr. The resultant solution which is acidic, requires 185 mL of 0.500 M NaOH for neutralization. Calculate the mass (in mg) of Zn(OH)2 that was originally added (problem 53 in Purdue test bank)
To calculate the mass of Zn(OH)2 that was originally added, we will follow these steps:
1. Determine the moles of HBr that were neutralized by NaOH:
The volume of HBr solution is given as 0.500 L, and the concentration is 0.50 M. Using the formula C = n/V (concentration = moles/volume), we can calculate the moles of HBr in the solution:
n(HBr) = C(HBr) x V(HBr)
n(HBr) = 0.50 mol/L x 0.500 L = 0.250 moles
2. Equate the moles of HBr neutralized to the moles of OH- from NaOH:
From the balanced chemical equation, we know that 1 mole of HBr reacts with 1 mole of NaOH. So, the moles of OH- from NaOH are also 0.250 moles.
3. Determine the moles of Zn(OH)2 that reacted:
From the balanced chemical equation, we know that 1 mole of Zn(OH)2 reacts with 2 moles of OH-. Therefore, the moles of Zn(OH)2 are half of the moles of OH-:
n(Zn(OH)2) = 0.250 moles / 2 = 0.125 moles
4. Calculate the mass of Zn(OH)2:
The molecular mass of Zn(OH)2 is given as 99.41 amu.
The molar mass of Zn(OH)2 can be calculated by summing the atomic masses of its constituents (1 Zn + 2 O + 2 H):
M(Zn(OH)2) = (1 x atomic mass of Zn) + (2 x atomic mass of O) + (2 x atomic mass of H)
M(Zn(OH)2) = 1 x 65.38 amu + 2 x 16.00 amu + 2 x 1.01 amu
M(Zn(OH)2) = 65.38 amu + 32.00 amu + 2.02 amu
M(Zn(OH)2) = 99.40 amu (approximately)
Now, we can calculate the mass of Zn(OH)2:
Mass(Zn(OH)2) = n(Zn(OH)2) x M(Zn(OH)2)
Mass(Zn(OH)2) = 0.125 moles x 99.40 g/mol
Mass(Zn(OH)2) = 12.425 g (approximately)
5. Convert the mass to milligrams:
We know that 1 g = 1000 mg, so the mass in milligrams is:
Mass(Zn(OH)2) = 12.425 g x 1000 mg/g
Mass(Zn(OH)2) = 12,425 mg
Therefore, the mass of Zn(OH)2 that was originally added is approximately 12,425 mg.
The reaction is:
Zn(OH)2 + 2HBr ---> ZnBr2 + 2H2O
This reaction shows that moles of HBr = (1/2) moles of HBr.
Moles of HBr = (liters)(moles/L) or
moles HBr = (0.500 L)(0.50 mol/L) = _____?
moles of ZnBr2 = (1/2) moles of HBr
Multiply the moles of ZnBr2 by its formula mass to get the mass of ZnBr2 in grams. Convert that to milligrams.