Suppose you have a 0.200 M solution of the nitrogen-containing weak base NX. Suppose you wish to titrate 25 mL of this 0.200 M NX solution with a 0.100 M solution of the strong acid HNO3. Given that the Kb value of NX is 6.50x10-5, complete each of the following:
a) What volume of HNO3 solution (in mL) will be required to reach the equivalence point of this titration?
b) What will be the pH of the solution at this point?
c) What will be the pH of the solution after addition of ½ of the required HNO3?
d) What will be the pH of the solution after addition of 75.0 mL of HNO3?
I see your posts as about the same although some may be strong acid/strong base while others may be strong acid/weak base (or the reverse). After you understand how to do these you can work them yourself. Pick ONE of the posts and clarify what you don't understand. I don't mind helping your through it but I don't want to work all of them for you. Each point on the titration curve contains a mixture of the reactants. Key to these problems is to recognize what you have and where you are on the titration curve.
To answer each part of this question, you will need to use the concept of titration and the properties of weak bases and strong acids. Here's how you can approach each part:
a) What volume of HNO3 solution (in mL) will be required to reach the equivalence point of this titration?
The equivalence point of a titration is reached when the moles of acid equal the moles of base. To find the volume needed at the equivalence point, you can use the formula:
(V1)(C1) = (V2)(C2)
Where:
V1 = volume of the base (0.025 L)
C1 = concentration of the base (0.200 M)
V2 = volume of the acid (unknown)
C2 = concentration of the acid (0.100 M)
Rearranging the formula, we get:
V2 = (V1)(C1) / C2
= (0.025 L)(0.200 M) / 0.100 M
= 0.050 L or 50 mL
Therefore, 50 mL of the HNO3 solution will be required to reach the equivalence point.
b) What will be the pH of the solution at this point?
At the equivalence point of an acid-base titration, the solution contains only the salt formed from the neutralization reaction. In this case, it will be the salt formed from the reaction between the weak base NX and the strong acid HNO3.
Since HNO3 is a strong acid, it fully dissociates in water. The resulting solution will contain the conjugate acid of NX and the anion from the strong acid. These ions will not undergo any further hydrolysis, so the resulting solution will be neutral, with a pH of 7.
c) What will be the pH of the solution after addition of ½ of the required HNO3?
Since we are adding half of the required HNO3 for neutralization, it means we have added 25 mL of the 50 mL needed to reach the equivalence point.
To find the moles of HNO3 added, we can use the formula:
n = C x V
Where:
n = moles
C = concentration
V = volume
moles of HNO3 added = (0.100 M)(0.025 L)
= 0.0025 mol
Since HNO3 is a strong acid, it fully dissociates in water, and the moles of HNO3 added will also be equal to the moles of H+ produced.
To find the concentration of H+ in the solution, we need to consider the initial volume and concentration of the base NX, as well as the volume and moles of HNO3 added.
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, A- represents the conjugate base of NX, which will be formed by the reaction between the added HNO3 and NX.
Given that Kb = 6.50x10^-5, we can calculate the pKa using the equation:
Kw = Ka x Kb
Where Kw is the ionization constant of water (1.0 x 10^-14) and Ka is the ionization constant of the weak acid formed by the conjugate of NX.
1.0 x 10^-14 = Ka x 6.50 x 10^-5
Solving for Ka:
Ka = 1.0 x 10^-14 / (6.50 x 10^-5)
The pKa can be found by taking the negative logarithm base 10 of Ka:
pKa = - log Ka
Using the pKa, you can plug it into the Henderson-Hasselbalch equation along with the concentrations of A- and HA to find the pH of the solution after adding half of the required HNO3.
d) What will be the pH of the solution after addition of 75.0 mL of HNO3?
To find the pH after adding 75.0 mL of HNO3, you need to determine the moles of HNO3 added, calculate the concentrations of H+ and A-, and use the Henderson-Hasselbalch equation as explained above to find the pH.