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3 years ago

How does the end point differ from the equivalence point of a titration?

Physics

1 answer:

Gwar [14]3 years ago

3 0

Answer:

Equivalence point and end point are terminologies in pH titrations and they are not the same. 

Explanation:

In a titration the substance added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the acid and base titrated determines the nature of the final solution.

At equivalence point the number of moles of the acid will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the pH at equivalence point. 

A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating neutral nature of the solution. 

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luda_lava [24]

Answer:

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2 years ago

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nlexa [21]

Answer: 2859.78 k

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3 years ago

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serg [7]

It accelerates in speed and also in change in direction.

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3 years ago

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Dvinal [7]

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2 years ago

Two identical conducting spheres, having charges of opposite sign, attract each otherwith a force of 0.108 n when separated by 5

lisov135 [29]

Let's start from the final situation. After the two spheres are connected with the conducting wire, the total charge distributes equally between the two spheres (because they are identical). We can call the charge on each sphere $Q/2$ , with Q being the total charge.
The electrostatic force in this situation is 0.0360 N, so we can write
$F=k \frac{( \frac{Q}{2} )^2}{r^2}$
where k is the Coulomb's constant and $r=50.0 cm=0.50 m$ is the separation between the two spheres. Using F=0.0360 N, we can find the value of Q, the total charge shared between the two spheres:
$Q= \sqrt{ \frac{4Fr^2}{k} } = \sqrt{ \frac{4(0.0360 N)(0.50 m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} }=2.0 \cdot 10^{-6}C$

Now let's go back to the initial situation, before the conducting wire was attached; in this situation, the two spheres have a charge of $q_1$ and $q_2$ , whose sum is Q:
$Q=q_1 + q_2$
The electrostatic force between the two spheres in the initial situation is:
$F=k \frac{q_1 q_2}{r^2}$
And since we know F=0.108 N, we find
$q_1 q_2 = \frac{Fr^2}{k} = \frac{(0.108 N)(0.50 m)^2}{8.99 \cdot 10^9 N m^2 C^{-2}}=3.0 \cdot 10^{-12} C$
But the problem tells us that the two spheres have charges of opposite sign, so we must put a negative sign:
$-3.0 \cdot 10^{-12} C = q_1 q_2$

So now we have basically a system of 2 equations:
$2.0 \cdot 10^{-6} C = Q = q_1 + q_2$
$-3.0 \cdot 10^{-12} C = q_1 q_2$
If we solve it, we find the initial charge on the two spheres:
$q_1 = -1 \cdot 10^{-6}C$
$q_2 = +3 \cdot 10^{-6 } C$

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3 years ago

How does the end point differ from the equivalence point of a titration?​

How does the end point differ from the equivalence point of a titration?