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marysya [2.9K]
3 years ago
5

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

on as present in blood plasma. Identify the acid and its conjugate base. Write the ionization equation and the Ka expression. 6.2 X 10^(8) a. The Ka of H2PO4 is 6.2 x 108. What is the pka? Is this acid weaker or stronger than H2CO3? Which buffer system is more optimal for regulating pH in the body? b. Using LeChatlier's Principle, what happens if the urine is acidified (H+ ions added)?
Chemistry
1 answer:
Katen [24]3 years ago
3 0

Answer:

The ionization equation is

H_{2}PO_{4}^{-}  +H_{2}O ⇄HPO_{4}^{-2}  +H_{3}O^{+} (1)

Explanation:

The ionization equation is

H_{2}PO_{4}^{-}  +H_{2}O ⇄HPO_{4}^{-2}  +H_{3}O^{+} (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}

The pKa is

-Log (Ka) = -Log (6.2x10^{-8}) = 7.2

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1)  moves to the left neutralizing the excess proton concentration.

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[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of FCH_{2}COOH

FCH_{2}COOH(aq) + H_{2}O (l)    FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = [H_{3}O^{+}] =[FCH_{2}COO^{-}] = 0.00770 M

pK_{a} = 2.59K_{a} = 10^{-2.59}   = 0.00257 M

K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}

0.00257 = \frac{0.00005929}{x - 0.00770}

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

Therefore, initial concentration of the weak acid is <u>0.031 M</u>

4 0
3 years ago
In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba(OH)2 was added to 60.0 mL of 0.600 M HCl. The reaction caused the temp
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Answer:

ΔH = 57.04 Kj/mole H₂O

Explanation:

60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)

=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)

=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)

=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat

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=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)

ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O

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3 years ago
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
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Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

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Answer:

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