CHEM 1212K Lecture Notes - Lecture 9: Stoichiometry, Weak Base, Methylamine

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5 Aug 2016

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Calculate the ka of an acid (or kb of a base) , given the concentration and the percent dissociation. Problem #1: a 0. 150 m solution of acetic acid (shorthand formula = hac) is found to be 1. 086% dissociated. The equation for the dissociation of hac is: [ac ] = [h+] = 1. 629 x 10 3 m: for the [hac], we use the concentration given. We could subtract 1. 629 x 10 3 from it, but this is usually not done. Warning: you might have a teacher who does ask you to do it. You might want to check to make sure: now, we insert into the ka expression and solve: x = [(1. 629 x 10 3) (1. 629 x 10 3)] / 0. 15 x = 1. 77 x 10 5. Note: in april 2009, when the chemteam solved the above example on paper, he wrote 0. 0186 rather than 0. 01086.

Related Questions

1) The pKa of acetic acid at 25 oC is 4.75. What is the pH of a 10-3 M solution: a) at 25 oC, b) at 50 oC? Show your work. (6 pts)

2) Which dissociation would proceed at 25 oC once the compound is added to water? Show your dissociation reactions. (12 pts)

a) NaHCO3 b) Mg(OH)2

3) Using analytical methods, determine the pH, pOH and concentration of all species for: a) 10-8 M HCl b) 10-3 M NaAc
Show all your equations and assumptions. (12 pts)

4) Using the graphical method (pC vs pH), find the pH, [HAc] and [Ac-] in a solution to which 10-3 moles of sodium acetate, NaAc, and 2 x 10-4 moles of acetic acid, HAc, pKa = 4.75, have been added per liter of solution. (25 pts)

5) Using the trial-and-error method, calculate the pH of a 10-3 M HAc following the addition of 400 mL of 10-3 M NaOH and 800 mL of 10-3 M NaOH. Estimate pH before attempting to calculate it exactly. (25 pts)

6) A 10-3 M buffer solution of NaHCO3/Na2CO3 is used to maintain pH = 10.
a) Can this solution be used for this purpose? Explain? (5 pts) -4 -4 b) What is the pH of the buffer solution after the addition of each: a) 10 M NaOH and b) 10 M HCl? (15 pts)

salmonminnow552

Acrylic acid, whose formula is HC₃H₃O₂, is used in the manufacture of plastics. A 0.77 aqueous solution of acrylic acid has a pH of 2.18. What is K_afor acrylic acid?

To solve this problem, first, we must formulate an ICE Table. The equation for the equilibrium of dissociation of chloroacetic acid is represented as below,


I	0.10	0	0
C	-x	x	x
E	0.10-x	x	x

We can solve for x by solving for the concentration of H+, given by the following formula:

This value is also equal to the concentration of CICH2CO2-. The concentration of the acid will be equal to 0.10 minus the value of x. From the given values, we may now solve for the Ka of the acid using the following formula:

Therefore, the value of Ka of acid is 1.82×10-3.

whitesardine29

CHEM 1212K Lecture Notes - Lecture 9: Stoichiometry, Weak Base, Methylamine

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