First the student pipetted 25.00ml of the cloudy ammonia solution into a 250.00ml conical flask. 59.00 ml of 0.100 mole per liter HCl(aq) was immediately added to the conical flask which reached with the ammonia in solution.

Question

The excess (unredacted) HCl was then titranted with 0.050M Na2CO3(aq). 21.50 ml of Na2CO3 (aq) was required.
Calculator the concentration of the ammonia in the cloudy ammonia solution?

Answer 1

To calculate the concentration of ammonia in the cloudy ammonia solution, we'll follow these steps:

1. Determine the moles of HCl added:
   The concentration of HCl is given as 0.100 mol/L, and the volume used is 59.00 mL.

   \[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.100 , \text{mol/L} \times 0.05900 , \text{L} = 0.00590 , \text{mol} \]

2. Determine the moles of Na2CO3 used:
   The concentration of Na2CO3 is given as 0.050 mol/L, and the volume used is 21.50 mL.

   \[ \text{Moles of Na2CO3} = \text{Concentration} \times \text{Volume} = 0.050 , \text{mol/L} \times 0.02150 , \text{L} = 0.001075 , \text{mol} \]

3. Determine the moles of excess HCl neutralized by Na2CO3:
   The reaction between Na2CO3 and HCl can be understood through the balanced equation: \[ \text{Na2CO3} + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H2O} + \text{CO2} \]

   From the equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the moles of HCl neutralized by the Na2CO3 is:

   \[ \text{Moles of HCl neutralized} = 2 \times \text{Moles of Na2CO3} = 2 \times 0.001075 , \text{mol} = 0.002150 , \text{mol} \]

4. Determine the moles of HCl that reacted with ammonia:
   The moles of HCl that were originally present (0.00590 mol) minus the moles that were neutralized by Na2CO3 gives us the moles of HCl that reacted with the ammonia:

   \[ \text{Moles of HCl reacted with NH3} = \text{Moles of HCl added} - \text{Moles of HCl neutralized} \]

   \[ \text{Moles of HCl reacted with NH3} = 0.00590 , \text{mol} - 0.002150 , \text{mol} = 0.00375 , \text{mol} \]

5. Determine the moles of ammonia:
   The reaction between ammonia (NH3) and HCl can also be summarized in a balanced equation: \[ \text{NH3} + \text{HCl} \rightarrow \text{NH4Cl} \]

   This indicates a 1:1 ratio between HCl and NH3. Therefore, the moles of NH3 that reacted is equal to the moles of HCl that reacted with it:

   \[ \text{Moles of NH3} = 0.00375 , \text{mol} \]

6. Concentration of ammonia in the original solution:
   The ammonia solution was initially dissolved in 25.00 mL (0.025 L). Therefore, the concentration of ammonia in that solution can be calculated as follows:

   \[ \text{Concentration of NH3} = \frac{\text{Moles of NH3}}{\text{Volume of solution (L)}} \]

   \[ \text{Concentration of NH3} = \frac{0.00375 , \text{mol}}{0.025 , \text{L}} = 0.150 , \text{mol/L} \]

Therefore, the concentration of ammonia in the cloudy ammonia solution is 0.150 mol/L.