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hammer [34]
1 year ago
14

calculate the ratio of [sodium acetate] to [acetic acid] in a buffer solution that has a ph of 4.25. the ka for acetic acid at 2

5 °c is 1.7 × 10−5. please choose the correct answer from the following choices, and then select the submit answer button. answer choices
Chemistry
1 answer:
const2013 [10]1 year ago
5 0

The ratio of [sodium acetate] to [acetic acid] in a buffer solution is 0.30.

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 4.25; pH of buffer solution

pKa = -log(1.7 × 10−5) = 4.77

4.25 = 4.77 + log(cs/ck)

log(cs/ck) =4.25 - 4.77 = -0.52

cs/ck = 10∧(-0.52) = 0.30; ratio of sodium acetate to acetic acid

A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.

Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.

More about sodium acetate: brainly.com/question/24671704

#SPJ4

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3 0
1 year ago
Regarding the Lyman series of hydrogen, calculate the frequency of the n = 5 line.
zvonat [6]
The Lyman series can be expressed in the formula <span><span>1/λ</span>=<span>RH</span><span>(1−<span>1/<span>n2</span></span>) where </span><span><span>RH</span>=1.0968×<span>107</span><span>m<span>−1</span></span>=<span><span>13.6eV</span><span>hc

</span></span></span></span>Where n is a natural number greater than or equal to 2 (i.e. n = 2,3,4,...). Therefore, the lines seen in the image above are the wavelengths corresponding to n=2 on the right, to n=∞on the left (there are infinitely many spectral lines, but they become very dense as they approach to n=∞<span> (Lyman limit), so only some of the first lines and the last one appear).
 The wavelengths (nm) in the Lyman series are all ultraviolet
:2 3 4 5 6 7 8 9 10 11
 Wavelength (nm) 121.6 102.6 97.3 95 93.8 93.1 92.6 92.3 92.1 91.9 91.18 (Lyman limit)
In your case for the n=5 line you have to replace "n" in the above formula for 5 and you should get a value of 95 x 10^-9 m for the wavelength. then you have to use the other equation that convert wavelength to frequency. </span>
4 0
2 years ago
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