2 litre of NH3 at 30°C and at 0.20 atmosphere is neutralized by 134 mL of a solution of H2SO4. Calculate normality of H2SO4
To calculate the normality of H2SO4, we first need to gather some important information from the problem:
1. The volume of NH3 solution: 2 liters
2. The temperature of NH3 solution: 30°C
3. The pressure of NH3 solution: 0.20 atmosphere
4. The volume of H2SO4 solution: 134 mL
Now let's break down the steps to calculate the normality of H2SO4:
Step 1: Convert the NH3 volume to liters:
2 liters of NH3 solution = 2000 mL of NH3 solution
Step 2: Convert the H2SO4 volume to liters:
134 mL of H2SO4 solution = 0.134 liters of H2SO4 solution
Step 3: Determine the molarity of NH3:
We'll need the Ideal Gas Law to find the molarity (M) of NH3:
PV = nRT
Where:
P = pressure of the NH3 solution (in atmospheres)
V = volume of the NH3 solution (in liters)
n = number of moles of NH3
R = ideal gas constant (0.0821 L * atm / mol * K)
T = temperature of the NH3 solution (in Kelvin)
First, convert the NH3 solution temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 30 + 273.15
T(K) = 303.15 K
Now we can rearrange the Ideal Gas Law equation to solve for moles (n) of NH3:
n = (PV) / (RT)
n = (0.20 atm) * (2000 mL / 1000 mL/L) / ((0.0821 L * atm / mol * K) * (303.15 K))
Step 4: Calculate the moles of NH3:
Solve the equation for n:
n = (0.20 atm) * (2 L) / (0.0821 L * atm / mol * K * 303.15 K)
Step 5: Determine the normality of H2SO4:
The normality (N) is defined as the number of equivalents (Eq) of solute per liter (L) of solution:
N = (Eq of solute) / (L of solution)
In this case, since 1 mole of H2SO4 is neutralized by 2 Eq of NH3, we can write:
Eq of H2SO4 = 2 * moles of NH3
Now, we can calculate the normality of H2SO4:
N of H2SO4 = (Eq of H2SO4) / (L of H2SO4 solution)
N of H2SO4 = (2 * moles of NH3) / (L of H2SO4 solution)
N of H2SO4 = (2 * n) / 0.134 L
By substituting the value of n calculated in Step 4, we can find the normality of H2SO4.
Use PV = nRT and solve for mols NH3. I get about 0.016 but you need to do it more accurately.
2NH3 + H2SO4 => (NH4)2SO4
N H2SO4 = # equivalents/L
#equivalents NH3 = 0.016
2 mols NH3 = 1 mol H2SO4
or 1 equivalent NH3 = 1/2 mol H2SO4
N H2SO4 = approx 0.016/2/0.134 = ?