Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq).
(a) What is the initial pH of the 0.20 M C6H5COOH(aq)?
(b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?
(c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?
mL
(d) Calculate the pH at the half stoichiometric point.
(e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?
mL
(f) Calculate the pH at the stoichiometric point.
Please show how to do it not just the answers.
The secret to this type problem is to know where you are on the titration curve.
C6H5COOH + KOH ==> C6H5COOK + H2O
For (a),
C6H5COOH + H2O ==> C6H5COO^- + H3O^+
Set up an ICE chart, substitute into Ka expression, and solve for H3O^+, then convert to pH.
For the others.
b,c,d. Calculate moles C6H5COOH initially. Calculate moles KOH added. Subtract moles KOH from moles benzoic acid to arrive at the excess benzoic acid, then substitute into either (1)Ka expression or (2) Henderson-Hasselbalch equation.
e. You find the stoichiometry point by mLa x Ma = mLb x Mb
a = acid; b = base.
f. You are at the equivalence point. The pH is determined by the hydrolysis of the salt, C6H5COOK (C6H5COO^- acting as a base).
C6H5COO^- + HOH ==> C6H5COOH + OH^-
Kb = (Kw/Ka) = (C6H5COOH)(OH^-)/(C6H5COO^-)
Post all of your work if you get stuck.
Set up ICE chart and solve for OH^- and convert to pH.
im still a bit confused about part c
i calculated:
n(C6H5COOH)initial= .006 mol
but when i calculate the KOH do i do the initial or with the added 15mL?
im a bit confused about that
To calculate the answers to the given questions, you will need to use the concept of acid-base titrations and the Henderson-Hasselbalch equation. Here is a step-by-step guide on how to solve each part:
(a) To find the initial pH of 0.20 M C6H5COOH(aq), we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
First, we need to find the pKa value for C6H5COOH. Assuming it is a weak acid, we can find the pKa value from reference sources or use the Ka value and take the negative logarithm (pKa = -log(Ka)).
Once you have the pKa value, substitute it into the Henderson-Hasselbalch equation along with the concentration of C6H5COOH ([HA]) to find the initial pH.
(b) After adding 15.0 mL (0.015 L) of 0.30 M KOH(aq) to C6H5COOH, we will have a new volume and concentration for C6H5COOH and KOH. To calculate the new concentration of C6H5COOH, use the equation:
M1V1 = M2V2
(0.20 M)(0.030 L) = M2(0.045 L)
Solve for M2, which will be the new concentration of C6H5COOH.
Now, use the Henderson-Hasselbalch equation as in part (a) to find the pH after the addition of KOH.
(c) To reach halfway to the stoichiometric point, we need to determine the volume of KOH required. The stoichiometric point occurs when the moles of acid are equal to the moles of base. From the balanced equation between C6H5COOH and KOH, determine the mole ratio.
Then, use the formula:
moles KOH = Molarity KOH × volume KOH
moles acid = Molarity acid × volume acid
Set the two equations equal to each other and solve for the volume of KOH.
(d) At the half stoichiometric point, we have equal moles of acid and base. Find the concentration of C6H5COOH at this point using the volume of KOH calculated in part (c). Then, substitute this concentration and the pKa value into the Henderson-Hasselbalch equation to calculate the pH.
(e) To reach the stoichiometric point, you need to determine the volume of KOH required. Again, use the stoichiometry from the balanced equation between C6H5COOH and KOH to determine the mole ratio.
Use the same formula as in part (c) to find the volume of KOH required.
(f) At the stoichiometric point, you have completely reacted all the C6H5COOH with KOH. Calculate the final concentration of the acid using the volume of KOH calculated in part (e). Once you have the concentration, substitute it and the pKa value into the Henderson-Hasselbalch equation to find the pH.
Note: Be sure to double-check your calculations and units throughout the process.