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Lorico [155]
3 years ago
5

Which one of the following conditions is met at the equivalence point of the titration of a monoprotic weak acid with a strong b

ase?A. The volume of strong base added from the buret must equal the volume of weak acid.B. The moles of strong base added must equal the moles of weak acid.C. pH
Chemistry
1 answer:
mrs_skeptik [129]3 years ago
4 0

Answer:

At the equivalence point of the titration of a monoprotic weak acid with a strong base: <em>B. the moles of strong base added must equal the moles of weak acid.</em>

Explanation:

In every titration, the equivalence point is defined as the point where the moles of the titrant and analyte are equal. For every acid-base titration, the equivalence point is defined as the point where the moles of the base is equal to the moles of the acid.

If the solutions of the acid and base are at a different concentration the volume added from the buret will not be the same as the volume of the analyte.

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Answer:

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Explanation:

Let's say the reaction is

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I like to consider heat as if it were a reactant or a product in a chemical equilibrium.

Another way to write the equilibrium would be

heat + R ⇌ 2P

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

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If you want to increase the amount of product, you increase the temperature. The system will try to get rid of the added heat by shifting to the right, thus forming more product.

(ii) Changing the pressure

If R and P are liquids or solids or in aqueous solution, changing the pressure will have no effect. Something must be in the gas phase for a change in pressure to affect the position of equilibrium.

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A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr
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Answer:

184.62 ml

Explanation:

Let p_1, v_1, and T_1 be the initial and p_2, v_2, and T_2 be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

p_1=p_2 ...(i)

v_1 = 200 ml

T_1= 26 ^{\circ}C = 273+26 =299 K

T_2= 3 ^{\circ}C = 273+3 =276 K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)

For the final condition,

p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)

Equating equation (i), and (ii)

\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}

\frac{v_1}{T_1}=\frac{v_2}{T_2}  [from equation (i)]

v_2=\frac{T_2}{T_1} \times v_1

Putting all the given values, we have

v_2=\frac{276}{299} \times 200 = 184.62 \; ml

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

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