Saccharin has a pKa of 2.32
It has a formula of HNC7H4SO3
If you dissolve 4.6 grams of saccharin in 478 ml of water, what is the resulting pH ?
Convert pKa to Ka. pKa - -log Ka
Let's call saccharin just HS.
M HS = grams/molar mass which I've estimated to be about 0.03 but you need a closer answer than that.
.....HS ==> H^+ + S^-
I..0.03.....0.....0
C....-x.....x.....x
E..0.03-x...x.....x
Substitute the E line into the Ka expression and solve for x = (H^+). Then pH = -log(H^+)
Note that I expect you will need to use the quadratic formula as I don't think 0.03-x will be close to 0.03.
To find the resulting pH of a solution containing saccharin, we need to determine if saccharin is an acid or a base and calculate the concentration of H+ ions or OH- ions in the solution.
Saccharin is a weak acid, and its acidity is determined by the dissociation of H+ ions. The pKa value of 2.32 indicates the acidity strength of saccharin.
First, we need to calculate the molar mass of saccharin (HNC7H4SO3):
H = 1 * 1 = 1
N = 14 * 1 = 14
C = 12 * 7 = 84
S = 32 * 1 = 32
O = 16 * 3 = 48
Total molar mass = 1 + 14 + 84 + 32 + 48 = 179 g/mol
Next, we can convert the mass of saccharin (4.6 grams) into moles:
moles = mass / molar mass = 4.6 / 179 = 0.0256 mol
Given that the solution contains 478 mL of water, we can convert this volume to liters:
volume = 478 mL = 478 / 1000 = 0.478 L
Now, we can calculate the concentration of saccharin in the solution:
concentration = moles / volume = 0.0256 mol / 0.478 L = 0.0536 M
Since saccharin is a weak acid, it undergoes partial dissociation in water. The resulting H+ concentration can be calculated using the acid dissociation constant (Ka) equation:
Ka = [H+][A-] / [HA]
The dissociation of saccharin can be represented as follows:
HNC7H4SO3 ⇌ H+ + NC7H4SO3-
[H+] is the concentration we want to find.
The equilibrium constant expression (Ka) is defined as:
Ka = [H+][NC7H4SO3-] / [HNC7H4SO3]
As saccharin is a weak acid, we can approximate that at equilibrium, the concentration of [NC7H4SO3-] and [H+] will be approximately equal.
Simplifying the equation:
Ka = [H+] * [HNC7H4SO3] / [HNC7H4SO3]
Ka = [H+] (since the concentrations of [HNC7H4SO3] and [H+] are approximately equal.)
Now, we need to take the logarithm of both sides to solve for [H+]:
log Ka = log [H+]
pKa = -log [H+]
Rearranging the equation:
-log [H+] = pKa
[H+] = 10^(-pKa)
Plugging in the given pKa value:
[H+] = 10^(-2.32)
Using a calculator, we find:
[H+] = 0.004786 M
Since [H+] = [A-] (approximately), the pH can be calculated using the equation:
pH = -log [H+] = -log (0.004786) ≈ 2.32
Therefore, the resulting pH of the solution containing 4.6 grams of saccharin dissolved in 478 mL of water is approximately 2.32.