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jarptica [38.1K]
3 years ago
10

Determine the dissociation constants for the following acids. Express the answers in proper scientific notation where appropriat

e
a. acid a has a pka of 6.0 what is its ka?
b. acid b has a pka of 8.60 whats its ka?
c. acid c has a pka of -2.0 whats its ka?

How do you determine?
Chemistry
1 answer:
Tems11 [23]3 years ago
8 0

Explanation:

Ka is the acid dissociation constant. pKa is simply the -log of this constant.

Mathematically;

pKa = - log Ka

Ka = 10^-pKa

a. acid a has a pka of 6.0 what is its ka?

Ka = 10^-6 = 10⁻⁶

b. acid b has a pka of 8.60 whats its ka?

Ka = 10^-8.6 = 2.51e⁻⁹

c. acid c has a pka of -2.0 whats its ka?

Ka = 10^-(-2) = 10^2 = 100

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Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

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Hydration of alkynes gives good yields of single compounds only with symmetrical or terminal alkynes. Draw the major organic pro
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Answer:

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