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Arte-miy333 [17]
1 year ago
14

the titration of a polyprotic acid with sufficiently different pka 's displays two equivalence points. why?

Chemistry
1 answer:
morpeh [17]1 year ago
4 0

If Ka1 and Ka2 are sufficiently distinct and a polyprotic acid is titrated with a strong base, the pH curve will contain two equivalence points because the two acidic protons will be titrated sequentially.

<h3>What is titration?</h3>

Titration is a method of chemical analysis in which the amount of a sample's constituents is determined by adding a precise amount of a different substance—one that the desired constituent reacts with in a specific, known proportion—to the sample being tested.

The titrating reagent, or titrant, is gradually added to the process using a standard solution (i.e., a solution of known concentration), which is administered using a burette, which is essentially a long, graduated measurement tube with a stopcock and a delivery tube at its bottom end. The addition comes to an end at the comparable point.

To know more about titration, visit:

brainly.com/question/2728613
#SPJ4

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3 0
2 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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