At 25oC, the equilibrium pressure of ammonia vapour above a 0.500 M aqueous ammonia solution is 0.84 kPa. Calculate the Henry's law constant for ammonia in M/bar. Do not include units. Include 4 significant figures.
so, you have 1/2 mole, per liter, at .84kPa.
so change kPa to bar, .84kpa*.001bar/100Pa=840*1e-5=.0084bar
constant(M/p)=.5/.0084=59.52
ok thankyou
To calculate the Henry's law constant for ammonia in M/bar, we can use the equation:
Henry's law constant (K) = (equilibrium pressure of ammonia) / (concentration of ammonia)
Given:
Equilibrium pressure of ammonia (P) = 0.84 kPa
Concentration of ammonia (C) = 0.500 M
First, we need to convert the pressure from kilopascals (kPa) to bars (bar):
1 kPa = 0.01 bar
Pressure (P) in bars = (0.84 kPa) * (0.01 bar/kPa) = 0.0084 bar
Now, we can calculate the Henry's law constant:
Henry's law constant (K) = (0.0084 bar) / (0.500 M)
Henry's law constant (K) ≈ 0.0168 bar/M
Therefore, the Henry's law constant for ammonia in M/bar is approximately 0.0168 (to 4 significant figures).
To calculate the Henry's law constant for ammonia in M/bar, we can use the equation:
C = k * P
where:
C is the concentration of ammonia in M (molarity),
k is the Henry's law constant in M/bar,
P is the partial pressure of ammonia in bar.
Given:
C = 0.500 M (molarity of the aqueous ammonia solution),
P = 0.84 kPa = 0.84/1000 bar (conversion of pressure from kPa to bar).
We need to solve for k.
Rearranging the equation, we get:
k = C / P
Substituting the given values:
k = 0.500 M / (0.84/1000 bar)
Simplifying:
k = 0.500 M / 0.00084 bar = 595.24 M/bar
Therefore, the Henry's law constant for ammonia is 595.2 M/bar.