You have 100.7 g of saccharin, a sugar substitute, and you want to prepare a pH = 5.80 solution. What volume of solution can be prepared? For saccharin (HC7H4NSO3), pKa = 11.70 (pKa = −log Ka).
To calculate the volume of solution that can be prepared, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
In this case, we want to prepare a solution with a pH of 5.80 and we have saccharin (HC7H4NSO3) with a pKa of 11.70.
From the equation, we can rearrange it to solve for the ratio of [A-] to [HA]:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given values into the equation, we have:
[A-]/[HA] = 10^(5.80 - 11.70)
[A-]/[HA] = 10^(-5.90)
[A-]/[HA] = 1.26 x 10^(-6)
Now, let's assume that the amount of saccharin used in the solution is x grams. Since the molar mass of saccharin (HC7H4NSO3) is 183.2 g/mol, we can calculate the moles of saccharin:
moles of saccharin = x / 183.2
From the balanced equation of saccharin, we can see that 1 mol of saccharin produces 1 mol of A- and 1 mol of HA. This means that the moles of A- and HA will be the same as the moles of saccharin:
moles of A- = moles of HA = x / 183.2
Since the ratio of [A-] to [HA] is 1.26 x 10^(-6), we can write:
[A-] / [HA] = (x / 183.2) / (x / 183.2) = 1.26 x 10^(-6)
Simplifying the equation:
1 = 1.26 x 10^(-6)
This implies that the volume of solution that can be prepared is infinite. However, this is not possible since we only have 100.7 g of saccharin.
Therefore, we cannot prepare a solution with pH = 5.80 using the given saccharin (HC7H4NSO3).
To calculate the volume of solution that can be prepared, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa value of the acidic component (in this case, saccharin) and the ratio of its conjugate base (C7H4NSO3-) to the acid (HC7H4NSO3).
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([C7H4NSO3-] / [HC7H4NSO3])
We are given the pKa value of saccharin as 11.70 and the desired pH as 5.80. Rearranging the equation, we have:
log([C7H4NSO3-] / [HC7H4NSO3]) = pH - pKa
Now, let's solve for the ratio of [C7H4NSO3-] / [HC7H4NSO3]:
[C7H4NSO3-] / [HC7H4NSO3] = 10^(pH - pKa)
Taking the antilog of both sides, we get:
[C7H4NSO3-] / [HC7H4NSO3] = antilog(pH - pKa)
In this case, the ratio [C7H4NSO3-] / [HC7H4NSO3] represents the conjugate base to acid ratio of saccharin, which is equal to the volume of the solution (in liters) that can be prepared.
Given that you have 100.7 g of saccharin, we can calculate the number of moles of saccharin using its molar mass. The molar mass of saccharin (HC7H4NSO3) is the sum of the atomic masses of each element in the compound. Upon calculating, we find that the molar mass is approximately 183.18 g/mol.
Number of moles of saccharin = Mass / Molar mass
Number of moles = 100.7 g / 183.18 g/mol
Now, divide the number of moles of saccharin by the ratio [C7H4NSO3-] / [HC7H4NSO3] to find the volume of the solution:
Volume of solution = Number of moles / [C7H4NSO3-] / [HC7H4NSO3]
Please substitute the appropriate values into the equation and calculate the result to find the volume of solution that can be prepared.