The side chain of histidine has a pKa of about 6. What percent of histidine side chains would be deprotonated at pH 7.5
1 answer:
The percent of histidine side chains would be deprotonated at pH 7.5 is 5.77 %.
<h3>What is pKa?</h3>The term pKa refers to the negative logarithm of the acid dissociation constant (Ka). The pH is the negative logarithm of the hydrogen ion concentration.
Hence;
Ka = Antilog (-6) =
[H^+] = Antilog (-7.5) =
We now have to use the formula;
α =
α =
α = 5.77 %
Hence, the percent of histidine side chains would be deprotonated at pH 7.5 is 5.77 %.
Learn more about percent dissociation: brainly.com/question/12273293
You might be interested in
Answer:
197mL of 0,506M HCl
Explanation:
The reaction of HCl + BaCO₃ is:
BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.
The moles of BaCO₃ in 9,85 g are:
9,85 g of BaCO₃ × = <em>0,0499 moles of BaCO₃</em>
As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:
0,0499 moles of BaCO₃ × =<em> 0,0998 moles of HCl</em>
If you have a 0,506M HCl, you need to add:
0,0998 moles of HCl× = 0,197 L ≡ 197mL
I hope it helps!
Answer: 3.0 moles of water are needed to react with a mole of 3.0 of Na
Explanation:
The balanced chemical reaction of sodium with water is as follows:
According to stoichiometry:
2 moles of sodium reacts with = 2 moles of water
Thus 3.0 moles of sodium reacts with =
moles of water
3.0 moles of water are needed to react with a mole of 3.0 of Na