Consider the titration of 50mL of 2.0M HNO3 with 1.0M KOH. At eah step, add the following: 0ml, 25ml, 50ml.
a) Write a reaction to show the initial reaction
b)ICE table
c) determine the spectators present after
d)Calculate the pH
Will you please explain, in detail, exactly what you don't understand about each step?
How do i write a reaction?
a) The initial reaction between HNO3 (nitric acid) and KOH (potassium hydroxide) can be represented as follows:
HNO3 + KOH -> KNO3 + H2O
b) To set up an ICE table, we need to analyze the changes in concentration of the relevant species present in the reaction. In this case, we are looking at the titration of 2.0M HNO3 with 1.0M KOH. Let's fill in the table for each step:
Step 1: 0 mL KOH added
HNO3 + 0 KOH -> KNO3 + H2O
Step 2: 25 mL KOH added
HNO3 + 25/1000 L KOH -> KNO3 + H2O
Step 3: 50 mL KOH added
HNO3 + 50/1000 L KOH -> KNO3 + H2O
Note: We convert milliliters (mL) to liters (L) by dividing by 1000.
c) To determine the spectator ions present after each step, we need to assess which ions remain unchanged throughout the reaction. In this case, the spectator ions would be the potassium ion (K+) from KOH and the nitrate ion (NO3-) from HNO3. These ions do not participate in the overall reaction; they are merely "spectators" and do not change.
d) To calculate the pH, we need to determine the concentration of H+ (hydronium ions) in the final solution. Since we are starting with a strong acid (HNO3) and titrating it with a strong base (KOH), we can assume that complete neutralization will occur. Therefore, at the equivalence point, the concentration of H+ will be equal to the concentration of OH- (hydroxide ions) before they react.
At the equivalence point, when 50 mL of 1.0M KOH is added to 50 mL of 2.0M HNO3, the number of moles of H+ and OH- will be equal. The reaction between them will form water (H2O). Thus, to determine the concentration of H+ in the solution, we can use the following equation:
[H+] = [OH-] = (moles OH-) / (total volume)
The total volume will be the sum of the volumes of HNO3 and KOH used. Using the given concentrations and volumes:
[H+] = [OH-] = (2.0 mol/L * 0.050 L) / (0.050 L + 0.050 L)
= 0.2 mol/L
To convert the H+ concentration to pH, we can use the formula:
pH = -log[H+]
Therefore, in this case, the pH would be:
pH = -log(0.2)
≈ 0.70
So, the pH of the solution after the titration is approximately 0.70.