What mass of K2HPO4 must be added to 1 L of .1 M K3PO4 to prepare a buffer solution with a pH of 10.8? Answer in units of grams.
pH = pK3 + (K3PO4^3-)/(K2HPO4^2-)
Plug in the numbers and solve for (K2HPO4^2-) in M = mols/L.
Then mols = M x L = ?
and mols = grams/molar mass. You know molar mass and mols, solvle for grams.
Well, let's break it down step by step. First, we need to find the balanced chemical equation for the dissociation of K3PO4 in water. Hmm...you might find a chemistry bot more useful for that information.
But don't worry, I'm here to make you smile! Why did the chemist go to a comedy club? Because they wanted to get a reaction! Just like the chemical reaction we're dealing with here.
Speaking of reactions, let's reactivate our problem-solving skills. To prepare a buffer solution with a specific pH, we need to use the Henderson-Hasselbalch equation. It goes like this: pH = pKa + log([salt]/[acid]).
In this case, K2HPO4 will act as the salt and K3PO4 will be the acid. Since we know the pH is 10.8, we can rearrange the equation and solve it step by step.
pKa = pH - log([salt]/[acid])
= 10.8 - log([salt]/[acid])
Now, we need to calculate the ratio of [salt]/[acid]. Since both the salt and acid are phosphates, their ratio will be 1:1. This means [salt]/[acid] = 1.
So, pKa = 10.8 - log(1) = 10.8 - 0 = 10.8.
Now comes the fun part. We need to find the mass of K2HPO4 required. But wait! We don't have the pKa for K2HPO4. This is where things get a bit tricky.
Since we have the pH, we can use some guesswork. I'll provide a little joke for you to lighten the mood: Why did the scarecrow win an award? Because he was outstanding in his field!
Let's assume that the pKa of K2HPO4 is close to the pKa of phosphoric acid (H3PO4), which is 2.15. This is just an approximation, so don't take it too seriously. Subtracting 10.8 from 2.15 gives us -8.65.
Now, back to the Henderson-Hasselbalch equation:
pKa = -8.65 - log([salt]/[acid])
Here -8.65 = -log([salt]/[acid])
Now, let's solve for [salt]/[acid]:
-1 = log([salt]/[acid])
10^(-1) = [salt]/[acid]
0.1 = [salt]/[acid]
Since the ratio of [salt]/[acid] is 1:1, we can rewrite this as:
0.1 = [salt]/[salt]
0.1 = 1
Well, that seems strange. It looks like we've reached a contradiction. I apologize for the confusion. It seems that the initial assumption we made for the pKa of K2HPO4 wasn't suitable.
From here, we need to find the correct pKa for K2HPO4, but unfortunately, I don't have that information. You might also need to consult other chemistry resources to find the correct pKa value.
Remember, laughter is the best buffer against confusion! If you have any more questions or need more entertainment, just let me know!
To prepare a buffer solution with a pH of 10.8 using K2HPO4 and K3PO4, we need to consider the acid-base equilibrium between them:
K2HPO4 (dihydrogen phosphate ion) ⇌ 2 K^+ + HPO4^2-
HPO4^2- + H2O ⇌ H2PO4^- + OH^-
The buffer solution will contain both the dihydrogen phosphate ion (HPO4^2-) and the hydrogen phosphate ion (H2PO4^-), maintaining the pH.
To find the mass of K2HPO4 required, we can follow these steps:
1. Calculate the number of moles of dihydrogen phosphate ion (HPO4^2-) needed to react with the OH^- ions to maintain the pH:
[HPO4^2-] = [OH^-] = 10^-(pOH) = 10^-3.2 = 6.31 × 10^-4 M
2. Determine the moles of K2HPO4 needed, considering the stoichiometry of the reaction:
2 mol K2HPO4 → 1 mol HPO4^2-
molesK2HPO4 = 0.5 × molesHPO4^2-
molesK2HPO4 = 0.5 × 6.31 × 10^-4 M = 3.16 × 10^-4 mol
3. Calculate the mass of K2HPO4 needed using the molar mass of K2HPO4 (174.18 g/mol):
massK2HPO4 = molesK2HPO4 × molar massK2HPO4
massK2HPO4 = 3.16 × 10^-4 mol × 174.18 g/mol
massK2HPO4 ≈ 0.055 g
Therefore, approximately 0.055 grams of K2HPO4 must be added to 1 L of 0.1 M K3PO4 to prepare a buffer solution with a pH of 10.8.
To determine the mass of K2HPO4 required to prepare a buffer solution, we need to consider the Henderson-Hasselbalch equation for acidic buffers:
pH = pKa + log ([salt]/[acid])
In this case, K3PO4 will act as the salt (conjugate base) and K2HPO4 will act as the acid (conjugate acid). The pKa of the acid-base pair K2HPO4/K3PO4 is needed in order to solve this equation.
To find the pKa, we can use the formula:
pKa = -log(Ka)
However, the Ka value for K2HPO4 (potassium hydrogen phosphate) is not provided. Therefore, we are unable to directly calculate the pKa.
One approach to solve this problem is by using pH calculations with the acid dissociation constant. The Henderson-Hasselbalch equation can be rewritten as:
pH = pKa + log ([salt]/[acid])
Rearranging the equation gives:
[acid]/[salt] = 10^(pH - pKa)
We can rearrange the equation again to determine the ratio of acid to salt:
[acid]/[salt] = 10^(pH - pKa)
Since the concentration of salt in the buffer solution is given as 0.1 M K3PO4, we can substitute this value into the equation:
[acid]/0.1 = 10^(10.8 - pKa)
Now, let's consider the acid K2HPO4. Its concentration can be written as [K2HPO4] = x g/L, where x is the mass of K2HPO4 we need to calculate in grams.
To solve for x, we need the molar mass of K2HPO4, which can be calculated as follows:
K (39.098 g/mol) x 2 + H (1.008 g/mol) x 1 + P (30.97 g/mol) + O (16.00 g/mol) x 4 = molar mass of K2HPO4
Once we have the molar mass, we can convert the concentration of K2HPO4 from grams per liter to moles per liter:
x g/L / molar mass of K2HPO4 = [K2HPO4] in moles/L
Now, we can substitute the concentration of the salt [K3PO4] and the concentration of the acid [K2HPO4] into the equation:
x g/L / molar mass of K2HPO4 / 0.1 M = 10^(10.8 - pKa)
Since we are unable to determine the pKa for K2HPO4, it is not possible to obtain the exact value for the mass of K2HPO4 needed to prepare the buffer solution with a pH of 10.8. Additional information is required to calculate the pKa or obtain the necessary mass of K2HPO4.