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Artemon [7]
3 years ago
9

How many moles of solid NaF would have to be added to 1.0 L of 0.115 M HF solution to achieve a buffer of pH 3.46? Assume there

is no volume change. Ka for HF = 7.2 × 10 –4
Chemistry
1 answer:
astra-53 [7]3 years ago
7 0

Answer : The  moles of solid NaF is, 1.09 mole.

Explanation : Given,

pH = 3.46

K_a=7.2\times 10^{-4}

Concentration of HF = 0.115 M

Volume of solution = 1.0 L

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (7.2\times 10^{-4})

pK_a=4-\log (7.2)

pK_a=3.14

Now we have to calculate the moles of HF.

\text{Moles of }HF=\text{Concentration of }HF\times \text{Volume of solution}

\text{Moles of }HF=0.115M\times 1.0L=0.115mol

Now we have to calculate the moles of NaF.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{\text{Moles of NaF}}{\text{Moles of HF}}

Now put all the given values in this expression, we get:

3.46=3.14+\log (\frac{\text{Moles of NaF}}{0.115})

\text{Moles of NaF}=1.09mol

Thus, the moles of solid NaF is, 1.09 mole.

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Which term best describes this reaction? 2 carbons double bonded to each other, with CH 3 above the left C and H above the right
Ierofanga [76]

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addition polymerization

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2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu; 83.79% of 52Cr, with an
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Answer:

Average atomic mass  = 51.9963 amu

Explanation:

Given data:

Abundance of Cr⁵⁰ with atomic mass= 4.34% ,  49.9460 amu

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Abundance of Cr⁵³ with atomic mass =9.50%,  52.9407 amu

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Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass +....n)  / 100

Average atomic mass  = (4.34×49.9460)+(83.79×51.9405) +(9.50×52.9407)+ (2.37×53.9389) / 100

Average atomic mass =  216.7656 + 4352.0945 + 502.9367 +127.8352 / 100

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