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A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL

snow_lady [41]

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

$pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})$

$pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43$

$pH = 14 - pOH = 14 - 4.43 = 9.57$

Now, the perchloric acid added will react with ammonia:

$n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles$

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O

$n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles$

Finally, we can calculate the pH of the resulting solution:

$pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})$

$pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98$

$pH = 14 - 4.98 = 9.02$

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!

5 0

3 years ago

Many important biochemicals are organic acids, such as pyruvic acid ( p K a = 2.50 ) and lactic acid ( p K a = 3.86 ) . The conj

Ivenika [448]

Answer:

Pyruvic acid: conjugate base

Lactic acid: conjugate base

Explanation:

The ratio of conjugate base to conjugate acid can be found using the Henderson-Hasselbalch equation when the pH and pKa are known.

pH = pKa + log([A⁻]/[HA])

The equation can be rearranged to solve for the ratio:

pH - pKa = log([A⁻]/[HA])

[A⁻]/[HA] = 10^(pH-pKa)

Now we can calculate the ratio for the pyruvic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 2.50) = 79433

[A⁻] = 79433[HA]

There is a much higher concentration of the conjugate base.

Similarly for lactic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 3.86) = 3467

[A⁻] = 3467[HA]

For lactic acid the conjugate base also dominates at pH 7.4

6 0

4 years ago

PLEASE ANSWER

romanna [79]

Answer: D

Explanation:

4 0

3 years ago

Problem PageQuestion The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium

Shalnov [3]

Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The reaction that takes place is:

2NaN₃(s) → 2Na(s) + 3N₂(g)

2. We use PV=nRT to calculate the moles of N₂ that were produced.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 K

n = 0.334 mol

Now we convert N₂ moles to NaN₃ moles:

0.334 mol N₂ * $\frac{2molNaN_{3}}{3molN_2}$ = 0.223 mol NaN₃

Finally we convert NaN₃ moles to grams, using its molar mass:

0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃

6 0

4 years ago

The enthalpy change of reaction is also called

Vikentia [17]

C) Heat of the reaction.

7 0

3 years ago

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