A student titrates HCl with 0.0125M KOH. Starting with 50.00 ml of the acid it takes 40.0 ml of base to reach equilavence point.
A. What is the concentration of the HCl?
B. What is the initial pH of the acid solution?
C. What is the pH of the reaction mixture after the addition of 20.00 ml of base?
D. What is the pH at equivlance point?
E. What is the pH after the addition of 50.00 ml of the base?
To answer the questions, we need to understand the concept and calculations involved in acid-base titration.
Acid-base titration is a process in which a known concentration of an acid is reacted with a known concentration of a base to determine the unknown concentration of the acid or base. In this case, we have a known concentration of KOH (0.0125M) and the volume of base needed to reach equivalence point (40.0 ml).
A. To find the concentration of HCl, we can use the principle of stoichiometry. Since the reaction between HCl and KOH is 1:1, the number of moles of KOH used is equal to the number of moles of HCl present in the initial volume.
Moles of KOH = concentration of KOH * volume of KOH = 0.0125 M * 40.0 ml = 0.50 mmol
Since the ratio is 1:1, the concentration of HCl is 0.50 mmol/50.00 ml = 0.0100 M.
B. To find the initial pH of the acid solution, we can use the formula for pH, which is defined as the negative logarithm of the hydrogen ion concentration. The initial concentration of HCl is 0.0100 M, so the initial hydrogen ion concentration is also 0.0100 M. Taking the negative logarithm, we get pH = -log(0.0100) = 2.00.
C. To find the pH of the reaction mixture after adding 20.00 ml of base, we need to first calculate the moles of base added. The moles of base added can be calculated using the volume and concentration of KOH added.
Moles of KOH added = concentration of KOH * volume of KOH added = 0.0125 M * 20.00 ml = 0.25 mmol
The total volume of the reaction mixture will be 50.00 ml (initial volume) + 20.00 ml (volume of base added) = 70.00 ml.
To calculate the concentration of the acid in the mixture after adding the base, we subtract the moles of base added from the initial moles of acid, then divide by the total volume of the mixture.
Moles of acid in mixture = initial moles of acid - moles of base added = 0.50 mmol - 0.25 mmol = 0.25 mmol
Concentration of acid in mixture = moles of acid in mixture / total volume of mixture = 0.25 mmol / 70.00 ml = 0.0036 M
Taking the negative logarithm, we find pH = -log(0.0036) ≈ 2.44.
D. At equivalence point, the moles of acid are equal to the moles of base added. Therefore, the concentration of the acid at equivalence point is the same as the concentration of the base used, which is 0.0125 M. Taking the negative logarithm, we find the pH = -log(0.0125) ≈ 1.90.
E. After adding 50.00 ml of the base, the total volume of the reaction mixture will be 50.00 ml (initial volume) + 50.00 ml (volume of base added) = 100.00 ml. To calculate the concentration of the acid in the mixture after adding the base, we subtract the moles of base added from the initial moles of acid, then divide by the total volume of the mixture.
Moles of acid in mixture = initial moles of acid - moles of base added = 0.50 mmol - 0.50 mmol = 0.00 mmol
Concentration of acid in mixture = moles of acid in mixture / total volume of mixture = 0.00 mmol / 100.00 ml = 0.000 M
Taking the negative logarithm of 0.000 M gives us undefined pH as the concentration is zero.
In summary:
A. The concentration of HCl is 0.0100 M.
B. The initial pH of the acid solution is 2.00.
C. The pH of the reaction mixture after adding 20.00 ml of base is 2.44.
D. The pH at equivalence point is 1.90.
E. The pH after adding 50.00 ml of the base is undefined (zero concentration).