Hi everyone,

I have been trying to figure out this titration equation all day and for some reason I am stumped.

Here is the question in full:
Potassium hydrogen phthalate, KHP, can be obtained in high purity and is used to determine the concentrations of solutions of strong bases. Strong bases react wiht the hydrogen phthalate ion as follows:
HP- (aq) + OH- = H2O (l) +p2- (aq)

The molar mass of KHP is 204.2 g/mol and Ka for the HP- ion is 3.1 x 10^-6.

The Questions are: (a) If a titration experiment begins with 0.4885 g of KHP and has a final volume of 100ml, what is the ph of the equivalence point?

(b) If the titration required 38.55ml of NaOH solution to reach the end point, what is the concentration of the NaOH solution?

Thanks for anyone's help and direction with this.

a.

mol KHP = 0.4885g/molar mass = ?
M = mols/0.1 L = approximately 0.024 but you need to do it more accurately.
At the equivalence point the pH is determined by the hydrolysis of the P^2- ion.
.........P^2- + HOH ==> HP^- + OH^-
I.......0.024M...........0.......0
C..........-x...........x........x
E.......0.024-x.........x........x

Kb for P^2- = (Kw/k2 for H2P) = (x)(x)/(0.024-x)
Substitute all of the numbers and solve for x = OH^-, then convert to pH.
b.
mols KHP = 0.4485/molar mass KHP
mols NaOH = mols KHP (the equation is 1:1)
M NaOH = mols NaOH/L NaOH. You know mols NaOH and L NaOH solve for M NaOH.

THANK YOU!

To solve these questions, we need to use some concepts from acid-base titration and stoichiometry. Let's go step by step:

(a) To find the pH at the equivalence point, we need to determine the concentration of OH- ions present in the solution.

Step 1: Calculate the number of moles of KHP:
Given mass of KHP = 0.4885 g
Molar mass of KHP = 204.2 g/mol

Number of moles of KHP = mass / molar mass
= 0.4885 g / 204.2 g/mol

Step 2: Calculate the number of moles of OH- ions:
From the balanced equation, we know that 1 mol of KHP reacts with 1 mol of OH- ions. Since the concentration of KHP and OH- at the equivalence point is the same, the number of moles of OH- ions is equal to the number of moles of KHP.

Number of moles of OH- ions = Number of moles of KHP

Step 3: Calculate the concentration of OH- ions:
The volume of the solution at the equivalence point is 100 mL, which is equivalent to 0.100 L.

Concentration of OH- ions = Number of moles of OH- ions / Volume of solution
= Number of moles of KHP / Volume of solution

Step 4: Calculate pOH and convert it to pH:
pOH = -log10(concentration of OH- ions)
pH = 14 - pOH

(b) To determine the concentration of the NaOH solution, we can use the stoichiometry of the balanced equation.

Step 1: Write the balanced equation for the reaction between KHP and NaOH:
KHP + NaOH = H2O + NaP

Step 2: Calculate the number of moles of KHP:
The number of moles of KHP is already calculated in step 1(a).

Step 3: Calculate the number of moles of NaOH:
Since the balanced equation shows that 1 mol of KHP reacts with 1 mol of NaOH, the number of moles of NaOH is also equal to the number of moles of KHP.

Number of moles of NaOH = Number of moles of KHP

Step 4: Calculate the concentration of NaOH:
The volume of NaOH solution used is 38.55 mL, which is equivalent to 0.03855 L.

Concentration of NaOH = Number of moles of NaOH / Volume of solution

I hope this helps you solve the problem. Let me know if you need further assistance!

To solve this titration problem, we need to understand the reaction that occurs between the potassium hydrogen phthalate (KHP) and the strong base (NaOH).

The given equation for the reaction is:

HP- (aq) + OH- = H2O (l) + p2- (aq)

Now, let's break down the two questions step by step.

(a) If a titration experiment begins with 0.4885 g of KHP and has a final volume of 100 ml, what is the pH of the equivalence point?

To find the pH at the equivalence point, we need to determine the concentration of the strong base (OH-) at this point. In a titration, at the equivalence point, the moles of acid (KHP) are equal to the moles of base (NaOH) added.

First, we need to convert the mass of KHP to moles:

Molar mass of KHP = 204.2 g/mol

moles of KHP = mass of KHP / molar mass of KHP = 0.4885 g / 204.2 g/mol = 0.00239 mol

Since KHP is a monoprotic acid, it reacts in a 1:1 ratio with the strong base. Therefore, at the equivalence point, the moles of OH- will be equal to the moles of KHP.

Volume of NaOH solution at the equivalence point = final volume of titration = 100 ml

Now, we need to calculate the molarity of the NaOH solution at the equivalence point:

Molarity of NaOH = moles of NaOH / volume of NaOH solution (in liters)

moles of NaOH = moles of KHP = 0.00239 mol

volume of NaOH solution (in liters) = 100 ml = 100/1000 L = 0.1 L

Molarity of NaOH = 0.00239 mol / 0.1 L = 0.0239 M

Since NaOH is a strong base, it completely dissociates in water to produce OH- ions.

So, the concentration of OH- ions at the equivalence point is 0.0239 M.

Now, to determine the pH at the equivalence point, we need to calculate the pOH and then convert it to pH using the equation pH = 14 - pOH:

pOH = -log[OH-] = -log(0.0239) ≈ 1.62

pH = 14 - pOH = 14 - 1.62 ≈ 12.38

Therefore, the pH at the equivalence point is approximately 12.38.

(b) If the titration required 38.55 ml of NaOH solution to reach the end point, what is the concentration of the NaOH solution?

At the end point of the titration, the volume of NaOH added is the same as the volume of KHP present in the solution.

volume of NaOH solution = volume at the end point = 38.55 ml = 38.55/1000 L = 0.03855 L

The moles of NaOH added at the end point can be calculated using the molarity of NaOH:

moles of NaOH = Molarity of NaOH x volume of NaOH solution (in liters)

moles of NaOH = 0.0239 M x 0.03855 L = 0.000923 mol

Since the reaction equation shows a 1:1 ratio between NaOH and KHP, the moles of NaOH added at the end point are equal to the moles of KHP in the solution:

moles of KHP = 0.000923 mol

Now, we need to find the concentration of the NaOH solution:

concentration of NaOH = moles of NaOH / volume of NaOH solution (in liters)

concentration of NaOH = 0.000923 mol / 0.03855 L ≈ 0.024 M

Therefore, the concentration of the NaOH solution is approximately 0.024 M.

I hope this helps! Let me know if you have any further questions.