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Ira Lisetskai [31]
3 years ago
9

You are making a 1L solution that you want to be buffered with ammonia and ammonium. However ammonia is a gas that is difficult

to dissolve directly in water with good precision. Instead you will start with ammonium chloride and add sodium hydroxide which will react with the ammonium to form ammonia. Which of the following combinations will give you the optimal buffered solution?
a. 54 grams ammonium chloride and 108 grams sodium hydroxide
b. 54 grams ammonium chloride and 54 grams sodium hydroxide
c. 54 grams ammonium chloride and 40 grams sodium hydroxide
d. 54 grams ammonium chloride and 10 grams sodium hydroxide
Chemistry
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

54 grams ammonium chloride and 40 grams sodium hydroxide

Explanation:

A buffer is a solution that contains either a weak acid and its salt or a weak base and its salt, the solution is resistant to changes in pH. This means that, a buffer is an aqueous solution of either a weak acid and its conjugate base or a weak base and its conjugate acid.

A Buffer is used to maintain a stable pH in a solution, buffers can neutralize small quantities of additional acid of base. For any buffer solution, there is always a working pH range and a set amount of acid or base that can be neutralized before the pH will change. The amount of acid or base that can be added to a buffer before changing its pH is called its buffer capacity.

A good buffer mixture is supposed to have about equal concentrations of its both components. It is a rule of thumb therefore, that a buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other component.

The implication of this is that the ammonium chloride and sodium hydroxide should be of approximately the same concentration. If the masses are dissolved as shown in the answer, then we will have 1molL-1 of each component of the buffer in accordance with the rule of thumb stated above.

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motikmotik

Answer:

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Explanation:

To convert from representative particles to moles, Avogadro's Number: 6.02*10²³, which tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance.

We can use it in a ratio.

\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Multiply by the given number of molecules.

3.1*10^{24} \ molecules \ H_2O*\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Flip the ratio so the molecules of water cancel out.

3.1*10^{24} \ molecules \ H_2O*\frac {1 \ mol \ H_2O}{6.02*10^{23} \ molecules \ H_2O}

3.1*10^{24} *\frac {1 \ mol \ H_2O}{6.02*10^{23} }

\frac {3.1*10^{24} \ mol \ H_2O}{6.02*10^{23} }

Divide.

5.14950166113 \ mol \ H_2O

The original number of molecules has 2 significant figures: 3 and 1, so our answer must have the same. For the number we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

5.1 \ mol \ H_2O

There are about 5.1 moles of water in 3.1*10²⁴ molecules of water.

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Answer:

                      %age Yield =   96 %

Explanation:

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Step 1: <u>Calculate moles of Ba(NO₃)₂:</u>

Moles  =  Mass / M.Mass

Moles  =  75.1 g / 261.33 g/mol

Moles  =  0.2873 moles of Ba(NO₃)₂

Step 2: <u>Find out moles of BaSO₄ formed:</u>

According to balance chemical equation,

                  1 mole of Ba(NO₃)₂ produced  =  1 mole of BaSO₄

So,

        0.2873 moles of Ba(NO₃)₂ will produce  =  X moles of BaSO₄

Solving for X,

                      X =  0.2873 mol × 1 mol / 1 mol

                       X =  0.2873 moles of BaSO₄

Step 3: Calculate Theoretical Mass of BaSO₄:

Mass  =  Moles × M.Mass

Mass  =  0.2873 mol × 233.38 g/mol

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Step 4: <u>Calculate %age Yield as:</u>

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Formula Used:

                   %age Yield  =  (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

                   %age Yield  =  (64.4 g ÷ 67.07 g) × 100

                   %age Yield =  96.01 % ≈ 96 %

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Your Welcome.

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