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miss Akunina [59]
2 years ago
9

You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is made of 0.05

0 M HOAc and 0.050 M NaOAc. How do the pHs of the buffered solutions compare
Chemistry
1 answer:
xeze [42]2 years ago
7 0

The pH of buffered solution is:

The two buffered solutions have 'equal' pH values.

Comparison between two buffer solutions:

To compare the pH of two different buffer solutions with different concentrations, Henderson-Hasselbalch equation is used.

pH = pKa + log \frac{Salt}{Acid}

pKa of HOAc = 4.76

In solution 1:

Concentration of salt= 5 M

Concentration of acid= 5 M

Putting the values in the equation,

pH = 4.76 + log \frac{5}{5}

pH = 4.76 + log1

pH = 4.76

In solution 2:

Concentration of salt= 0.050 M

Concentration of acid= 0.050 M

Putting the values in the equation,

pH = 4.76 + log\frac{0.050}{0.050}

pH = 4.76 + log1

pH = 4.76

As the ratio  log\frac{salt}{acid} is the same for the different solutions.

Therefore, the pH of the solution remains the same in both the buffer solutions even if the two solutions have different concentrations.

Learn more about buffer solutions here,

brainly.com/question/15991816

#SPJ4

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Nitric monoxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO₂), a dark brown gas. If 5.895 mol of NO is mixed with 2
Schach [20]

Answer:

Limiting reactant: O2

grams NO2 produced = 230.276 g NO2

grams of NO unused = 26.67 gNO

Explanation:

2NO + O2 --> 2NO2

Step 1: Determine the molar ratio NO:O2

molar ratio NO:O2 = 5.895: 2.503 = 2.35

stoichiometric molar ratio NO:O2 = 2:1

So, O2 is the limiting reactant.

Step2: Determine the grams of NO2:

?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2

Step 3: Determine the amount of excess reagent unreacted

moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted

moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted

mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted

8 0
3 years ago
how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *
Vitek1552 [10]

Answer:

The number of moles =

Moles=4.97\times 10^{-8}

The number of molecules =

Molecules = 2.99\times 10^{16}

Explanation:

Volume of the sphere is given by :

V=\frac{4}{3}\pi r^{3}

here, r = radius of the sphere

radius=\frac{diameter}{2}

radius=\frac{6.0}{2}

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

V=\frac{4}{3}\pi 0.03^{3}

V=\frac{4}{3}\pi 2.7\times 10^{-5}

V=1.13\times 10^{-4}dm^{3}

V=1.13\times 10^{-4}L   (1 L = 1 dm3)

Now, concentration "C"=

C=4.4\times 10^{-4}moles/liter  

The concentration is given by the formula :

C=\frac{moles}{Volume(L)}

This is also written as,

Moles = C\times Volume

Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}

Moles=4.97\times 10^{-8}moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n"  mole of substance contain =( n x Na )

N_{a}=6.022\times 10^{23}

Molecules =

Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}

Molecules = 2.99\times 10^{16} molecules

7 0
3 years ago
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