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Vadim26 [7]
2 years ago
10

What letter on the model titration curve corresponds to the point where ph equals the numerical value of pka for hpr? what speci

es of hpr are present at this point?
Chemistry
1 answer:
irakobra [83]2 years ago
4 0

Letter C on the model titration curve corresponds to the point where pH equals the numerical value of pKa for HPr

<h3>What is a titration curve?</h3>

A titration curve is a graph of the pH of a solution against increasing volumes of an acid or a base that is added to the solution.

The pH of a solution is the negative logarithm to base ten of the hydrogen ion concentration and is a measure of the acidity or alkalinity of the solution.

The pKa is the acid dissociation constant of an acid solution.

In a titration of a strong acid and strong base, the pH at equivalence point is equal to the pKa of the acid.

The equivalence point is the point when equal moles of acids and base has reacted.

In the given titration curve, pH = pKa at point C.

In conclusion, for a titration curve of strong acid and base, at equivalence point, pH is equal to pKa of acid.

Learn more about equivalence point at: brainly.com/question/23502649

#SPJ1

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When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
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The question is incomplete, the complete question is:

When 177. g of alanine (C_3H_7NO_2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9^oC lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2^oC lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\Delta T_f=i\times K_f\times m

OR

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

  • <u>When alanine is dissolved in mystery liquid X:</u>

\Delta T_f=5.9^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant

m_{solute} = Given mass of solute (alanine) = 177. g

M_{solute} = Molar mass of solute (alanine) = 89 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m

  • <u>When KBr is dissolved in mystery liquid X:</u>

\Delta T_f=7.2^oC

i = Vant Hoff factor = ?

K_f = freezing point depression constant = 2.37^oC/m

m_{solute} = Given mass of solute (KBr) = 177. g

M_{solute} = Molar mass of solute (KBr) = 119 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63

Hence, the van't Hoff factor for potassium bromide in X is 1.63

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The answer is B. 

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