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4 years ago

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.75.

Chemistry

1 answer:

lions [1.4K]4 years ago

5 0

Answer:

a) 0.0323 M

b) ΔpH = 0.10

Explanation:

A buffer is a solution of a weak acid and its conjugate base, or of a weak base and its conjugate acid. This solution keeps the pH almost constant because it has equilibrium between the acid and the base. So, when acid is added, it reacts with the base and when a base is added, it reacts with the acid.

To calculate the pH in a buffered solution, we may use the Henderson-Hasselbalch equation:

$pH = pKa + log\frac{[A^-]}{[HA]}$

Where pKa is the -logKa and Ka is the acid constant, and [A⁻] is the concentration of the base (or ionic compound of the acid), and [HA] is the concentration of the acid.

a) C₆H₅NH₂ is the base and its conjugate acid is C₆H₅NH₃⁺. pKa + pKb = pKw, where pKw is from the water equilibrium and is equal to 14.

pKa = 14 - 9.13 = 4.87

So,

pH = pKa + log [C₆H₅NH₂]/[C₆H₅NH₃⁺]

5.75 = 4.87 + log (0.245)/[C₆H₅NH₃⁺]

5.75 - 4.87 = log (0.245) - log [C₆H₅NH₃⁺]

log[C₆H₅NH₃⁺] = -0.6108 - 0.88

log [C₆H₅NH₃⁺] = -1.4908

[C₆H₅NH₃⁺] = $10^{-1.4908}$

[C₆H₅NH₃⁺] = 0.0323 M

b) First, let's calculated the concentration of NaOH, knowing that the molar masses are

MNa = 23 g/mol

MO = 16 g/mol

MH = 1 g/mol

MNaOH = 40 g/mol

n = m/M

n = 0.360/40 = 0.009 mol

The concentration is: n/V = 0.009/1.5 = 0.006 M.

The base reacts with the acid to form more base compound. Than the concnetration of aniline and aniline hydrochloride will be know:

[C₆H₅NH₂] = 0.245 + 0.006 = 0.2510 M

[C₆H₅NH₃⁺] = 0.0323 - 0.006 = 0.0263 M

So, the new pH will be:

pH = 4.87 + log(0.2510)/(0.0263)

pH = 4.87 + 0.98

pH = 5.85

ΔpH = 5.85 - 5.75

ΔpH = 0.10

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Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

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Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

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Answer:

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Explanation:

First we convert 9.5 g of NaCl into moles of NaCl, using its molar mass:

9.5 g ÷ 58.44 g/mol = 0.16 mol NaCl

In 0.16 moles of NaCl there are 0.16 moles of sodium as well.

We now convert 0.16 moles of sodium into grams, using sodium's molar mass:

0.16 mol * 23 g/mol = 3.68 g

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