How many grams of NaOH(Mr 40) must be added to 100 ml of 0.1M H3PO4 to create phosphate buffer, pH 7.0?
I know I'm using HH-equation. H3PO4, pka1=2.12, pka2=7.21, pka3=12.32.
Heres what I got:
pH=Pka+ log HPO4/H3PO4-HPO4
7.0=7.21 + log HPO4/.001-HPO4
is this right dr.Bob...I'm solving for HPO4, right?
Isn't a phosphate buffer @ pH about 7.2 composed of NaH2PO4 and Na2HPO4?
So you must add enough NaOH to completely neutralize the first hydrogen; therefore, you add 100mL x 0.1M mmoles NaOH to START. Then calculate the base(HPO4--/H2PO4-) you need to ADD to the NaOH initially used to arrive at the desired pH. .
100 ml * 0.1 M= 10 mmole
7.0 = 7.21 + log HPO4^-2/ 10- HPO4^-2
HPO4^-2 = 1.27M
how do i get to grams
Right, Dr.Bob?
Yes, you are correct. To create a phosphate buffer at pH 7.0, you need to solve for the concentration of HPO4 (dihydrogen phosphate ion).
The Henderson-Hasselbalch equation for this system is:
pH = pKa + log [HPO4] / [H2PO4-]
In this case, we consider [HPO4] as the concentration of dihydrogen phosphate ion, and [H2PO4-] as the concentration of monohydrogen phosphate ion.
So, let's go through the steps of solving for [HPO4]:
pH = 7.0
pKa = 7.21
Plugging these values into the Henderson-Hasselbalch equation:
7.0 = 7.21 + log [HPO4] / [H2PO4-]
Now, the concentration of [H2PO4-] can be calculated from the initial concentration of H3PO4 and its dissociation constant (pKa2 = 7.21).
Since H3PO4 is a triprotic acid, it will first lose a proton to become H2PO4-, and then lose another proton to become HPO4^2-. Therefore, at pH 7.0, the concentration of H2PO4- will be greater than that of HPO4^2-.
To calculate the concentration of [H2PO4-]:
[H2PO4-] = [H3PO4] * (10^-(pKa1 - pH))
Given:
[H3PO4] = 0.1 M
pKa1 = 2.12
pH = 7.0
[H2PO4-] = 0.1 * (10^-(2.12 - 7.0))
Now, with the concentration of [H2PO4-], we can substitute it back into the original equation to solve for [HPO4].
7.0 = 7.21 + log [HPO4] / (0.1 - [H2PO4-])
Simplifying the equation, you are solving for [HPO4].
Yes, you are on the right track. To create a phosphate buffer with a pH of 7.0, you need to use the Henderson-Hasselbalch equation. However, there is a small correction needed in your equation.
The Henderson-Hasselbalch equation for a phosphate buffer is:
pH = pKa + log([HPO4]/[H2PO4-])
Where:
- pH is the desired pH of the buffer (7.0 in this case)
- pKa is the ionization constant of the acid (H3PO4)
- [HPO4] is the concentration of the dihydrogen phosphate ion
- [H2PO4-] is the concentration of the monohydrogen phosphate ion
From the ionization constants you provided, pKa2 is the relevant value for the H2PO4- / HPO4 2- equilibrium.
Now, let's solve for [HPO4]:
7.0 = 7.21 + log([HPO4]/[H2PO4-])
To simplify the equation, we need to convert the concentrations from molarity (M) to moles per liter (mol/L). Therefore, [H2PO4-] would be 0.1 M (given by the initial concentration of H3PO4) and [HPO4] (which we are solving for) would be x grams of Na2HPO4.
Calculating the ratio of [HPO4]/[H2PO4-] requires knowing the molar masses of HPO4 and H2PO4-. The molar mass of HPO4 is 98 g/mol (2*1 + 31 + 16*4 = 98) and that of H2PO4- is 97 g/mol (1 + 31 + 16*4 = 97).
Substituting the values into the equation:
7.0 = 7.21 + log((x grams / 98 g/mol) / (0.1 M / 97 g/mol))
Simplifying further:
7.0 = 7.21 + log((x/98) / (0.1/97))
Now, you can solve this equation for x, which will give you the number of grams of Na2HPO4 needed to create the phosphate buffer with a pH of 7.0.
1. Subtract 7.21 from both sides:
7.0 - 7.21 = log((x/98) / (0.1/97))
-0.21 = log((x/98) / (0.1/97))
2. Take the antilog (exponentiate) both sides of the equation:
10^(-0.21) = (x/98) / (0.1/97)
3. Rearrange the equation to solve for x:
x = 10^(-0.21) * (0.1/97) * 98
Using a scientific calculator, evaluate the right-hand side of the equation to find the value of x. This will give you the number of grams of Na2HPO4 needed to create the phosphate buffer with a pH of 7.0.