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aniked [119]
3 years ago
13

What are the primary chemical components present in a phosphate buffer at ph 7.4? H3po4 and po43–?

Chemistry
1 answer:
fenix001 [56]3 years ago
8 0

Answer :

The correct answer for primary component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .

<u>Buffer solution :</u>

It is a solution of mixture of weak acid and its conjugate base OR weak base and its conjugate acid . It resist any change in solution when small amount of strong acid or base is added .

<u>Capacity of a good buffer : </u>

A good buffer is identified when pH = pKa .

From Hasselbalch - Henderson equation which is as follows :

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

If [A⁻] = [HA] ,

pH = pka + log 1

pH = pKa

This determines that if  concentration of  weak acid  and its conjugate base are changed in small quantity , the capacity of  buffer to maintain a constant pH is greatest at pka .  If the amount of [A⁻] or [HA] is changed in large amount , the log value deviates more than +/-  1M and hence pH .

Hence Buffer has best capacity at pH = pka .

<u>Phosphate Buffer : </u>

Phosphate may have three types of acid-base pairs at different pka ( shown in image ).

Since the question is asking the pH = 7.4

At pH = 7.4 , the best phosphate buffer will have pka near to 7.4 .

If image is checked the acid - base pair  " H₂PO₄⁻ and HPO₄²⁻  has pka 7.2 which is near to pH = 7.4 .

Hence we can say  , the primary chemical component of phosphate buffer at pH = 7.4  is H₂PO₄⁻ and HPO₄²⁻  .


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3 years ago
determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)
sasho [114]

The  mass  of water formed  is


<u><em>calculation</em></u>

Use  the  ideal   gas  equation   to  calculate the  moles of  NH3  and O2

that  is  Pv= n RT

where;  P= pressure,  

V=  volume,

n = number  of  moles,

R=gas   constant  = 0.0821  l .atm/ mol.K

make n the formula of  the subject  by diving   both side  by  RT

n =  PV /RT

The   moles of NH3

n= (1.50 atm  x 12.5 L) /(  0.0821 L. atm /mol.k   x 298 K)  =0.766  moles

The  moles  of  O2

=(1.1 atm  x 18.9  L) /  (  0.0821 L. atm/ mol.k   x 323 K) = 0.784  moles


write the reaction  between  NH3  and  O2

4 NH3  + 5 O2  →4 No  +6H2O


from  equation above  0.766  moles of NH3  reacted to produce  

0.766 x 6/4 =1.149 moles of H2O


0.784  moles of O2   reacted to  produce  0.784  x 6/5=0.9408  moles  of H20


since  O2  is totally  consumed, O2  is the limiting  reagent  and therefore  the  moles of H2O  produced=  0.9408  moles


mass  of  H2O  = moles x molar mass

 from  periodic table the  molar mass  of H2O  =  (1 x2)+16= 18  g/mol

mass = 18 g/mol  x 0.9408  moles= 16.93  grams


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