How do you get 1.8×10^-5? 16- and Weak Bases 713 initial concns: changes: equil

Question

How do you get 1.8×10^-5?

16- and Weak Bases 713 initial concns: changes: equil concns: (0.100- x) M 0.100 M x M x M Ka [CH3COOH] 0.100-x-1.8% 10% g the quadratic formula, but let's instead make a simplifying assumption lve this equation by usin uld so hat is often valid. Assume that r is very small compared with 0,100. That is, assume that (0.100-) 0.100 x2-0.100 × 1.8 × 10-5 = 1.8 × 10-6 for and x = [H3O+-V1.8 × 10-6 = 13x 10-3 M Now, we must check our assumption : 0.100-0.0013 0.099 0.100. Our assumption is good to about 1 part per 100 (1%) and is valid for a calculation involving two or three significant figures. Finally pH =-log[H3O+] =-log(1.3 × 10-3) =-(-2.89) = 2.89 Assess We observe that our answer is very close to the number on the pH meter heretore, our assumption to simplify the calculation was reasonable. This type

Explanation / Answer

1.8*10^-5 is the dissociation constant of acetic acid.

It will be given in the question some times or since It is a constant that can be obtained from chart of dissociation constants of various acids and bases.

Or you can also obtain it by knowing the pka value of acetic acid I,e equal to 4.75

Pka =- Log (Ka)

Ka = antilog (-pka)

=antilog (-4.75)

= 1.77*10^-5. Approximately 1.8*10-5

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