What is the pH of a solution of 40.0 ml of 0.800 M HCl titrated with 4.00 ml of 1.00 M Hydroxyl amine?
mols HCl = M x L = 0.800 x 0.040 = 0.032
mols HONH2 = M x L = 1.00 x 0.004 = 0.004
So you have 0.032-0.004 = 0.028 mols HCl remaining and the volume is 40 + 4 = 44 mL or 0.044 L.
So M HCl = mols/L and convert that to pH.
To determine the pH of the solution after the titration, we need to understand the chemical reaction that occurs between HCl and hydroxylamine (NH2OH).
The balanced chemical equation for the reaction between HCl and NH2OH is:
NH2OH + HCl -> NH3OH+ + Cl-
First, we need to determine the moles of HCl and NH2OH used in the reaction.
Molarity (M) is defined as moles of solute per liter of solution. To calculate the moles of a solution, we use the formula:
Moles = Molarity * Volume (in liters)
Given:
Volume of HCl solution = 40.0 ml = 0.040 L
Molarity of HCl solution = 0.800 M
Moles of HCl = 0.800 M * 0.040 L = 0.032 moles
Volume of NH2OH solution = 4.00 ml = 0.004 L
Molarity of NH2OH solution = 1.00 M
Moles of NH2OH = 1.00 M * 0.004 L = 0.004 moles
Since the balanced chemical equation is 1:1 between HCl and NH2OH, the moles of HCl and NH2OH are equal. Therefore, the amount of HCl used in the reaction is limited by the amount of NH2OH available.
Now, we can calculate the remaining moles of HCl after the reaction:
Remaining moles of HCl = Initial moles of HCl - Moles of NH2OH used
= 0.032 moles - 0.004 moles
= 0.028 moles
To calculate the concentration of HCl after the reaction, we need to divide the remaining moles by the new volume of the solution.
New volume of the solution = Volume of HCl solution + Volume of NH2OH solution
= 0.040 L + 0.004 L
= 0.044 L
Concentration of HCl = Remaining moles of HCl / New volume of the solution
= 0.028 moles / 0.044 L
≈ 0.6364 M
To find the pH, we can use the equation:
pH = -log10(H+ concentration)
Since HCl ionizes completely in water, the concentration of H+ ions is equal to the concentration of the HCl solution.
pH = -log10(0.6364)
≈ 0.197
Therefore, the pH of the solution after the titration is approximately 0.197.