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Svetach [21]
3 years ago
13

(a) carbonate buffers are important in regulating the ph of blood at 7.40. what is the concentration ratio of co2 (usually writt

en h2co3) to hco3- in blood at ph = 7.40? h2co3(aq) equilibrium reaction arrow hco3-(aq) + h+(aq) ka = 4.3 ✕ 10-7 webassign will check your answer for the correct number of significant figures. (b) phosphate buffers are important in regulating the ph of intracellular fluids at ph values generally between 7.1 and 7.2. what is the concentration ratio of h2po4- to hpo42- in intracellular fluid at ph = 7.15? h2po4-(aq) equilibrium reaction arrow hpo42-(aq) + h+(aq) ka = 6.2 ✕ 10-8 webassign will check your answer for the correct number of significant figures. (c) why is a buffer composed of h3po4 and h2po4- ineffective in buffering the ph of intracellular fluid? h3po4(aq) equilibrium reaction arrow h2po4-(aq) + h+(aq) ka = 7.5 ✕ 10-3
Chemistry
1 answer:
never [62]3 years ago
6 0
A) when the balanced equation of the reaction is:
H2CO3(aq) → HCO3 -(aq) + (H+)
and when we have Ka = 4.3 x10^-7 & PH = 7.4 
So first we will get PKa = -㏒ Ka
PKa = -㏒(4.3x10^-7) = 6.37 by substitution with Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
PH= PKa + ㏒[HCO3-]/[H2CO3]
㏒[HCO3-]/[H2CO3] = PH-Pka
[HCO3-] /[H2CO3] = 10^(7.4 - 6.37)
∴[HCO3-]/[H2CO3] = 11.7
∴[H2CO3]/[HCO3-] = 1/11.7 =  0.09

B) when The balanced equation for this reaction is:
H2PO42-(aq) → HPO4-(aq) + H+
and when we have Ka = 6.2x10^-8 & PH = 7.15
So Pka= -㏒Ka = -㏒(6.2x10^-8) = 7.2 by substitution by Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
7.15= 7.2 + ㏒[HPO4]/[H2PO4] 
-0.05 = ㏒[HPO4]/[H2PO4]
∴[HPO4]/[H2PO4] = 10^-0.05 = 0.89
∴[H2PO4]/[HPO4] = 1/0.89 = 1.12

c) H3PO4(aq) ↔ H2PO-(aq) + H+
the answer is: because we have Ka =7.5x10^-3 and it is a high value of Ka to make a good buffer, also we need a week acid with th salt of the week acid as H3PO4 is a strong acid so it does'nt make a goof buffer.


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7 0
3 years ago
Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.
mars1129 [50]

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

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Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

7 0
3 years ago
Barium nitrate and sodium sulfate solutions can be used to precipitate barium sulfate. How many grams
-Dominant- [34]

Answer:

40.8g of sodium sulfate must be added

Explanation:

The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:

Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)

That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:

<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>

75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄

<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>

0.287 moles * (142.04g / mol) =

<h3>40.8g of sodium sulfate must be added</h3>
7 0
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5 0
2 years ago
2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l) The ΔH for this reaction is 54.8 kJ . How much energy would be absorbed if
irakobra [83]
1) Chemical equation

<span>2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l)

2) Stoichiometric ratios

2 mol NH4Cl(s) : 54.8 KJ

3) Convert 24.7 g of NH4Cl into number of moles, using the molar mass

molar mass of NH4Cl = 14 g/mol + 4*1 g/mol + 35.5 g/mol = 53.5 g/mol

number of moles = mass in grams / molar mass

number of moles = 24.7 g / 53.5 g/mol = 0.462 moles

4) Use proportions:

2 moles NH4Cl / 54.8 kJ = 0.462 moles / x

=> x = 0.462 moles * 54.8 kJ / 2 moles = 12.7 kJ

Answer: 12.7 kJ
</span>
7 0
3 years ago
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