Answer:
a) 1.1974 M
b) 2.07%
Explanation:
In this case, we have a neutralization reaction between the HCl and ammonia, the acid base reaction is as follow:
HCl + NH₃ --------> NH₄⁺ + Cl⁻
a) As we can see, we have a 1:1 mole ration between the acid and the base, therefore we can use the following expression to calculate concentration of the ammonia:
M₁V₁ = M₂V₂
Replacing the data in here, we can solve for the concentration of ammonia:
M₂ = M₁V₁ / V₂
M₂ = 0.311 * 38.50 / 10
M₂ = 1.1974 M
This is the concentration of the ammonia solution.
b) Now we want to know the mass percent, in this case, we need to calculate the moles of ammonia first, and then, turn it into mass using the molar mass given (17 g/mol):
n = 1.1974 * 0.010 = 0.011974 moles
m = 0.011974 * 17 = 0.2036 g
The solution has a density of 0.983 g/mL and we have 10 mL therefore, we should have of mass:
d = m/V
m = d * V
m = 0.983 * 10 = 9.83 g
Finally the mass percent will be:
% = 0.2036 / 9.83 * 100
% = 2.07%
