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Lena [83]
3 years ago
13

Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pKa of 6.0, calculate the ratio of con

jugate base to acid at a pH of 5.0.
Choice of Weak Acid for a Buffer Which of these com-pounds would be the best buffer at pH 5.0: formic acid (pKa 5 3.8), acetic acid (pKa 5 4.76), or ethylamine (pKa 5 9.0)? Briefly justify your answer.
Chemistry
1 answer:
Free_Kalibri [48]3 years ago
7 0

Explanation:

According to the Handerson equation,  

          pH = pK_{a} + log \frac{\text{salt}}{\text{acid}}

or,      pH = pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}

Putting the given values into the above equation as follows.

     pH = pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}

       5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

      log \frac{\text{conjugate base}}{\text{acid}} = -1.0

or,      \frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}

                            = 0.1

Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.

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One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo
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Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J

Stape 2: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J

The total work is the sum of the two steps

W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J

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