Answer:
a) 11.64 M
b) 43 mL
c) 1.7 kg
Explanation:
a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:
1.18*1000 = 1180 g.
The mass of HCl will be then:
mHCl = 1180*0.36 = 424.8 g
The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:
nHCl = 424.8/36.5 = 11.64 mol
The molarity is the number of moles divided by the volume in L:
Molarity = 11.64 M
b) To prepare a solution by dilution of a concentrated one, we can use the equation:
C1V1 = C2V2
Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:
11.64*V1 = 2.00*0.250
V1 = 0.0429 L ≅ 43 mL
c) The neutralization will happen by the equation:
HCl + NaHCO₃ → NaCl + CO₂ + H₂O
So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:
nHCl = 1.75*11.64 = 20.37 mol
The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:
m = 84*20.37 = 1,711.08 g
m = 1.7 kg
