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Inessa05 [86]
3 years ago
7

Calculate the pH value in each of the following solutions, given their [H3O+] concentrations.

Chemistry
1 answer:
FrozenT [24]3 years ago
4 0

Answer:

See Explanations ...

Explanation:

In general, pH is a 'p-factor' expression which, simply put, is a way to express very small numbers (i.e.; exponential data with 10⁻ⁿ value ranges) in a more convenient form. That is, by definition, pX = -log(X) where X is the data value of interest. In practical terms, p-factor analysis can be applied to a number of physical & chemical measurements such as ...

pH => measure of acidity of solution = -log[H₃O⁺]

pOH => measure of alkalinity of solution = -log[OH⁻]

pKa => measure of weak acid ionization in aqueous solution = -log(Ka)

pKb => measure of weak base ionization in aqueous solution = -log(Kb)

pKsp => measure of salt ionization in aqueous solution = -log(Ksp)

Such can be applied to ranges of small-number values defining other chemical and physical properties.

For this problem:

Gastric Juice: [H₃O⁺] = 1.6 x 10⁻²M => pH = -log(1.6 x 10⁻²) = -(-1.80) = 1.8

Cow's Milk:  [H₃O⁺] = 2.5 x 10⁻⁷M => pH = -log(2.5 x 10⁻⁷) = -(-6.60) = 6.60

Tomato Juice:  [H₃O⁺] = 5.0 x 10⁻⁵M => pH = -log(5.0 x 10⁻⁵) = -(-4.30) = 4.30

Other Applications:

Given:

[OH⁻] = 6.30 x 10⁻¹³M => pOH = -log(6.30 x 10⁻¹³) = -(-12.2) = 12.2

Ka = 4.5 x 10⁻⁵ => pKa = -log(4.5 x 10⁻⁵) = -(-4.35) = 4.35

Kb = 8.2 x 10⁻⁶ => pKb = -log(8.2 x 10⁻⁶) = -(-5.09) = 5.09

Ksp = 5.5 x 10⁻¹⁰ => pKsp = -log(-5.5 x 10⁻¹⁰) = -(-9.26) = 9.26

Note: The values for Ka, Kb & Ksp are typically provided in tables of weak acid ionization constants (Ka-values), weak base ionization constants (Kb-values) or solubility product constants of salts (Ksp-values).

Hope this helps, Doc :-)

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<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

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The average rate of disappearance of ozone ... is found to  

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This means (ignoring time)

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Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

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Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
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<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

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K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

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