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vampirchik [111]
3 years ago
5

Which of the following buffers will be most effective at pH 9.25? Group of answer choices a mixture of 1.0 M HC2H3O2 and 1.0 M N

aC2H3O2 (Ka for acetic acid = 1.8 x 10-5) a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10) a mixture of 1.0 M HCl and 1.0 M NaCl a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 1
Chemistry
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

Explanation:

Step 1: Data given

pH of a buffer = pKa + log ([A-]/[Ha])

a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)

pH = -log( 1.8 * 10^-5) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 4.74

a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)

pH = -log( 4.9 * 10^-10) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 9.30

a mixture of 1.0 M HCl and 1.0 M NaCl

The solution made from NaCl and HCl will NOT act as a buffer.

HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.

a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)

Ka * Kb = 1*10^-14

Ka = 10^-14 / 1.76*10^-5

Ka = 5.68*10^-10

pH = -log( 5.68*10^-10) + log (1/1)

pH = -log( 5.68*10^-10)

pH = 9.25

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

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Gold is currently trading at very high price. Suppose that gold is selling for around $1860/ounce. How
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Answer:

The answer is "3.81041978"

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\to 1 \ OZ= 28.349523125 \ grams\\\\

              =28.349523125\times {1000} \ miligrams\\\\= 28349. 5231  \ miligrams\\

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
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Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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