Answer:
a. 0.182
b. 1.009
c. 1.819
Explanation:
Henderson-Hasselbach equation is:
pH = pKa + log [salt / acid]
Let's replace the formula by the given values.
a. 3 = 3.74 + log [salt / acid]
3 - 3.74 = log [salt / acid]
-0.74 = log [salt / acid]
10⁻⁰'⁷⁴ = 0.182
b. 3.744 = 3.74 + log [salt / acid]
3.744 - 3.74 = log [salt / acid]
0.004 = log [salt / acid]
10⁰'⁰⁰⁴ = 1.009
c. 4 = 3.74 + log [salt / acid]
4 - 3.74 = log [salt / acid]
0.26 = log [salt / acid]
10⁰'²⁶ = 1.819
Answer:
C.) HOCl Ka=3.5x10^-8
Explanation:
In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below
we Know that
pKa= -log(Ka)
therefore
A) pKa of HClO2 = -log(1.2 x 10^-2)
=1.9208
B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644
C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45
D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979
If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution
The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.
So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.
Hence, HOCl will be chosen for buffer construction.
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B.
Explanation:
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Explanation:
Haemoglobin consists of heme unit which is comprised of an <u>
</u> and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.
The prosthetic group of hemoglobin and myoglobin is - <u>Heme</u>
The organic ring component of heme is - <u>Porphyrin</u>
Under normal conditions, the central atom of heme is - <u>
</u>
In <u>deoxyhemoglobin</u> , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.
The central atom has <u>six</u> bonds: <u>four</u> to nitrogen atoms in the porphyrin, one to a <u>histidine</u> residue, and one to oxygen.