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Raul3140
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I need Help Please! Chemistry gurus please help!?
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!
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!
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!
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