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3 years ago

For which type of titration will the ph be acidic at the equivalence point?

Chemistry

2 answers:

ANTONII [103]3 years ago

8 0

It is
b. strong acid vs. weak base.

Ira Lisetskai [31]3 years ago

5 0

Equivalence point in an acid base titration is the point when there are equal moles of acid and base in the solution.

The pH at the equivalence point of a strong acid and a strong base will be 7 as the solution will be neutral. The conjugate acid and the conjugate base of a strong base and strong acid will be neutral.

$HCl(aq) + NaOH(aq)---> NaCl(aq) + H_{2}O(l)$

The pH at the equivalence point of a titration of weak acid and a strong base will be slightly basic as the conjugate base of the weak acid formed in the titration will be basic.

$CH_{3}COOH (aq) + NaOH (aq)--> CH_{3}COONa(aq) + H_{2}O(l)$

The pH at the equivalence point of a titration of strong acid and a weak base will be slightly acidic because the conjugate acid of the weak base is slightly acidic.

$HCl(aq) + NH_{3}(aq) ---> NH_{4}^{+}(aq) + Cl^{-}(aq)$

The pH at the equivalence point of the titration of a weak acid and a weak base is 7. But the pH changes are not sharp as both the acid and base are weak.

$CH_{3}COOH (aq)+NH_{3}(aq) -->CH_{3}COONH_{4}(aq)$

Therefore, the correct answer is B: Strong acid versus weak base

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4 years ago

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3 years ago

Read 2 more answers

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.

ruslelena [56]

Answer:

a) 0.168 moles O2

b) 9.50 grams O2

c) 0.01662 kg NO

d)88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen = 6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = 0.168 moles O2

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol = 9.50 grams O2

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = 0.01662 kg NO

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

Calculate moles of NH3 = 2.51 grams / 17.03 g/mol = 0.147 moles

Calculate moles of O2 = 3.76 grams / 32 g/mol = 0.118 moles

Determine the limiting reactant

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

Calculate moles H2O: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

Calculate mass H2O = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

Calculate % yield = (2.27/2.552)*100 % = 88.9 %

4 0

3 years ago