What Are the Steps to Solve These Chemistry Problems?

[Raul3140]

I need Help Please! Chemistry gurus please help!?

Homework Statement

I need Help Please! Chemistry gurus please help!?
1) 5.61 mL of 0.49M Pyridine, C5H5N, are titrated with 0.20 M HCl. What is the pH of the solution 2.86 mL after the equivalent point?
Kb (C5H5N)=1.8x10-9
Report the pH to 2 decimal places

2) The solubility of M2X3 (as M3+ and X2-) in water at 298K is 6.03 x 10(-3) M. Calculate Ksp for M2X3.

3) A 0.166 M NaBr and 0.048 M NaCl mixture is going to be treated with AgNO3. Calculate the % of bromide ion present when the choride starts to precipitate. Ksp (AgCl) = 1.8x10-10; (AgBr) = 5.0x10-13
Report your answer to two significant figures.

I tried and cannot get the correct answers...please help!

Homework Equations

The Attempt at a Solution

For 1) I got Ka from Kb by using 1.0E-14 and then I found out how much HCL was needed to reach the equivalence point and got
.2635 L by mult. and using Pyridine. but IDK how to get the HCL needed after addidtion of 2.86 ml. I used the same format as getting H3O and dividing it by total volume with 2.86 ml added and got 4.87 PH which is wrong!

For2) K_sp = [M3+]^2*[X2-]^3

ow Molarity of M3+ = 6.03 x 10(-3) M2X3*(2 mol M3+/1 mol M2X3) = 6.14E-13 mol = Molarity

IDK where to go after this

I tried solubility laws but I keep getting the wrong answer!

 

[Borek]

Show how you got the wrong answer.

 

[Raul3140]

I updated my attempt there.

 

[Borek]

First - the only role pyridine plays is it consumes HCl before endpoint. Final pH is a function of excess HCl only.

Second - what are concentrations of M³⁺ and X^2- if each liter of solution contains 6.03x10^-3 moles of M₂X₃?

 

[DeeNos]

Please help!Hello,

First of all, I am not a guru, but I am happy to help with your chemistry questions.

1) To find the pH at 2.86 mL after the equivalence point, we first need to calculate the moles of HCl added at the equivalence point. We know that at the equivalence point, the moles of HCl added is equal to the moles of pyridine present in the original solution. So, we can use the formula M1V1 = M2V2 to calculate the moles of HCl added at the equivalence point.
M1 = 0.20 M (concentration of HCl)
V1 = 5.61 mL (volume of pyridine solution)
M2 = M1 (since we are using the same concentration of HCl)
V2 = 2.86 mL (volume of HCl added at the equivalence point)
Solving for M1, we get M1 = 0.20 * 5.61 / 2.86 = 0.39 M.
Therefore, the moles of HCl added at the equivalence point is 0.39 mol.
Now, we can use the Henderson-Hasselbalch equation to calculate the pH at 2.86 mL after the equivalence point.
pH = pKa + log ([base]/[acid])
pKa = -log(Kb) = -log(1.8 * 10^-9) = 8.74
[base] = moles of pyridine remaining in the solution = 0.49 mol - 0.39 mol = 0.10 mol
[acid] = moles of HCl added at 2.86 mL after equivalence point = 0.39 mol
Substituting in the values, we get pH = 8.74 + log (0.10/0.39) = 8.74 - 0.41 = 8.33.
Therefore, the pH at 2.86 mL after the equivalence point is 8.33.

2) To calculate Ksp, we need to use the solubility product expression, which is Ksp = [M3+]^2 * [X2-]^3.
We know that the solubility of M2X3 is 6.03 * 10^-3 M. This means that the concentration of both M3