Question

An enzyme-catalyzed reaction is carried out in a solution containing 0.2 M tris((hydroxymethyl)aminomethane (known as Tris, pK, 8.1). Shown below is the chemical structure in its protonated form. *NH, HOCH, C- CH, OH CHLOH The pH of the reaction mixture at the start is 7.8. As a result of the reaction, 0.033 moles/liter of His consumed. What will be the ratio of Tris (free base) to Tris (protonated form) at the beginning of the reaction? A B. C. D. 5:1. 2:1. 1:1. 1:2. 5:1. Based on the information from question 6, what will be the ratio of Tris (free base) to Tris (protonated form) at the end of the reaction? 4:1. B. 2:1. 1:1. 1:2. 1:4.

Answer 1

Ans 1: It is given that the pKa value for Tris is 8.1 so using the relation pKa+pKb=14 we can calculate that pKb= 7.9. At the beginning of the reaction the pH is 7.8 and we know that

pH= pKa- log[HA]/[A^-]; putting the respective values we get 7.8= 8.1- log[HA]/[A^-].

0.3= log [HA]/ [A^-]; 10^(-0.3) = [A^-]/ [HA]; [A^-]/ [HA] = 1:2.

So the ratio of free base to the protonated form is 1:2. Option D is correct

Ans 2: As per the reaction 0.033 moles/liter of H⁺ is consumed this means that there is a decrease in the protonated form and increase in the basic form. SInce we know the ratio of basic to protonated form is 1:2 and we have the initial concentration 0.2 M, so the concentration of protonated form is 0.134 and the basic form is 0.066. When 0.033 moles/liter of H⁺ consumes the same amount is increased in basic form and decrease in the protonated form as the protonated form is breaking into free base and H⁺. The new concentration becomes 0.101 for protonated form and 0.099 for free base. Now taking the ratio we get 0.099/0.101 which is approximately equal to 1. So after the reaction completion the ratio of free base to the protonated form is 1:1. Option C is correct.