A25.0mL sample of 0.1MNH3 is titrated with 0.1M HCl.
A . calculate the pH value of the solution
after the following volume of 0.1M
HCl are added. 0.00 mL,5.00
mL,10.00 mL,15.00 mL,20.00
mL,22.00 mL,24.00
mL,25.00 mL and 26.00 mL
B. draw a table and put the pH value
corresponding to each value in ttable
C. Draw the titration curve
D. Name the appropriate indicator for this titration.
A. 0.00 mL: pH = 11.00
5.00 mL: pH = 9.00
10.00 mL: pH = 7.00
15.00 mL: pH = 5.00
20.00 mL: pH = 4.00
22.00 mL: pH = 3.50
24.00 mL: pH = 3.00
25.00 mL: pH = 2.50
26.00 mL: pH = 2.00
B. Volume of HCl (mL) | pH
0.00 | 11.00
5.00 | 9.00
10.00 | 7.00
15.00 | 5.00
20.00 | 4.00
22.00 | 3.50
24.00 | 3.00
25.00 | 2.50
26.00 | 2.00
C. The titration curve would look like a bell-shaped curve, with the pH increasing from 11.00 to 2.00 as the volume of HCl increases.
D. The appropriate indicator for this titration is phenolphthalein.
A. To calculate the pH value of the solution after the given volumes of 0.1M HCl are added, we need to consider the stoichiometry of the reaction between NH3 and HCl. The reaction can be represented as follows:
NH3 + HCl → NH4+ + Cl-
Using the balanced equation, we can calculate the moles of NH3 in the 25.0mL sample.
0.1M NH3 = 0.1 moles/L
Moles of NH3 = 0.1 moles/L * 0.025 L = 0.0025 moles
Now, let's determine the number of moles of HCl present in each volume of HCl added:
0.00 mL HCl: No HCl added yet, so moles of HCl remain 0.
5.00 mL HCl: 0.1M * (5.00 mL / 1000 mL) = 0.0005 moles HCl
10.00 mL HCl: 0.1M * (10.00 mL / 1000 mL) = 0.0010 moles HCl
15.00 mL HCl: 0.1M * (15.00 mL / 1000 mL) = 0.0015 moles HCl
20.00 mL HCl: 0.1M * (20.00 mL / 1000 mL) = 0.0020 moles HCl
22.00 mL HCl: 0.1M * (22.00 mL / 1000 mL) = 0.0022 moles HCl
24.00 mL HCl: 0.1M * (24.00 mL / 1000 mL) = 0.0024 moles HCl
25.00 mL HCl: 0.1M * (25.00 mL / 1000 mL) = 0.0025 moles HCl
26.00 mL HCl: 0.1M * (26.00 mL / 1000 mL) = 0.0026 moles HCl
To determine the pH of the solution at each volume of HCl added, we need to calculate the concentration of NH4+ and calculate the pOH and then pH.
B. pH values corresponding to each volume can be calculated using the concentration of NH4+ and the formula pH = 14 - pOH.
The concentration of NH4+ is equal to the moles of NH4+ divided by the total volume of the solution.
C. To draw the titration curve, plot the volume of HCl added on the x-axis and the pH value on the y-axis.
D. A suitable indicator for this titration is Methyl orange.
To calculate the pH value of the solution after each volume of 0.1M HCl is added, we need to take into account the reaction between NH3 (ammonia) and HCl (hydrochloric acid). NH3 reacts with HCl in a 1:1 ratio to form NH4+ (ammonium) and Cl- (chloride). The ammonium ion, NH4+, is acidic and contributes to the overall pH of the solution.
A. Calculating the pH value:
1. Before any HCl is added:
Initial volume of NH3 = 25.0 mL
Concentration of NH3 = 0.1 M
Number of moles of NH3 = Volume x Concentration = 25.0 mL x 0.1 M = 2.50 mmol
Total volume of the solution = 25.0 mL
To calculate the pH, we need to determine the concentration of NH4+ and OH- in the solution. At the beginning, there are only NH3 molecules present, and no NH4+ or OH- ions. Therefore, the solution is basic.
2. After each volume of HCl is added:
We need to consider the reaction between NH3 and HCl in a 1:1 ratio to determine how many moles of NH3 and HCl react.
NH3(aq) + HCl(aq) -> NH4+(aq) + Cl-(aq)
By using the stoichiometry of the reaction, for every mole of NH3 reacted, it will consume one mole of HCl. Therefore, the number of moles of NH3 remaining will decrease by the same amount as the moles of HCl added.
To calculate the pH, we need to determine the new concentration of NH3 and NH4+ after each addition of HCl.
B. pH values table:
Based on the above explanation, fill in the table with the volume of HCl added and the corresponding pH value calculated at each point.
Volume of HCl (mL) | pH value
-------------------|---------
0.00 | pH before any HCl is added
5.00 | pH after 5.00 mL of HCl is added
10.00 | pH after 10.00 mL of HCl is added
15.00 | pH after 15.00 mL of HCl is added
20.00 | pH after 20.00 mL of HCl is added
22.00 | pH after 22.00 mL of HCl is added
24.00 | pH after 24.00 mL of HCl is added
25.00 | pH after 25.00 mL of HCl is added
26.00 | pH after 26.00 mL of HCl is added
C. Titration curve:
To draw the titration curve, plot the pH values on the y-axis and the volume of HCl added on the x-axis. Connect the points to create a curve. The curve will show the changes in pH as the volume of HCl increases from 0.00 mL to 26.00 mL.
D. Indicator for the titration:
The appropriate indicator for this titration should have a pH range that overlaps with the pH range of the equivalence point. In the case of NH3 and HCl, the equivalence point occurs when equal moles of NH3 and HCl have reacted. The pH at this point will be approximately 7 (neutral).
A suitable indicator for this titration could be Phenolphthalein, which has a pH range of approximately 8.2 - 10.0. It changes color from colorless to pink as the pH of the solution increases above around 8.2.
Note: It's always important to consult your laboratory guidelines or a professional chemist when conducting experiments, as they can provide specific insights and safety measures.