If you mix 30 mL of 0.05 M K2HPO4 with 24 mL of 0.05 M KH2PO4, what % of the PO4 buffer is in the K2HPO4 form?
I'm just not sure at all how to start for this question because I thought K2HPO4 and KH2PO4 are both buffers?
Yes, you are correct that both K2HPO4 and KH2PO4 can act as buffers. However, in this question, we are looking to determine the percentage of the PO4 buffer that exists in the K2HPO4 form.
To do this, we need to calculate the moles of both K2HPO4 and KH2PO4 in the given solutions and then compare their amounts.
First, let's calculate the moles of K2HPO4:
molarity = moles/volume
moles of K2HPO4 = molarity × volume
moles of K2HPO4 = 0.05 M × 30 mL = 0.05 mol/L × 0.030 L = 0.0015 mol
Next, let's calculate the moles of KH2PO4:
moles of KH2PO4 = molarity × volume
moles of KH2PO4 = 0.05 M × 24 mL = 0.05 mol/L × 0.024 L = 0.0012 mol
Now, let's compare the moles of K2HPO4 and KH2PO4:
Total moles of PO4 buffer = moles of K2HPO4 + moles of KH2PO4
Total moles of PO4 buffer = 0.0015 mol + 0.0012 mol = 0.0027 mol
Now, let's determine the percentage of the PO4 buffer in the K2HPO4 form:
Percentage of K2HPO4 = (moles of K2HPO4 / total moles of PO4 buffer) × 100
Percentage of K2HPO4 = (0.0015 mol / 0.0027 mol) × 100 = 55.56%
Therefore, approximately 55.56% of the PO4 buffer is in the K2HPO4 form.
To determine the percentage of the PO4 buffer in the K2HPO4 form, you need to consider the dissociation of the two compounds in solution. K2HPO4 and KH2PO4 are indeed both buffer solutions, but they have different acid-base properties.
K2HPO4 is a diprotic acid, meaning it can donate two protons (H+) to a solution. In water, it dissociates into two K+ ions and one HPO4²- ion:
K2HPO4 ⇌ 2K+ + HPO4²-
On the other hand, KH2PO4 is a monoprotic acid, meaning it can donate only one proton (H+) to a solution. It dissociates into one K+ ion and one H2PO4- ion:
KH2PO4 ⇌ K+ + H2PO4-
To determine the percentage of the PO4 buffer in the K2HPO4 form, you first need to calculate the molar amounts (moles) of K2HPO4 and KH2PO4 in the given volumes and concentrations.
For K2HPO4:
Moles of K2HPO4 = Volume (in liters) × Concentration (M) = 0.030 L × 0.05 M = 0.0015 mol
For KH2PO4:
Moles of KH2PO4 = Volume (in liters) × Concentration (M) = 0.024 L × 0.05 M = 0.0012 mol
Next, you need to compare the stoichiometry of K2HPO4 and KH2PO4 in terms of the PO4²- ion. From the equations above, you can see that one K2HPO4 molecule produces one PO4²- ion, while one KH2PO4 molecule also produces one PO4²- ion.
Therefore, the total moles of PO4²- available in the mixture are:
Moles of PO4²- = Moles of K2HPO4 + Moles of KH2PO4 = 0.0015 mol + 0.0012 mol = 0.0027 mol
Finally, you can calculate the percentage of the PO4 buffer in the K2HPO4 form:
% of PO4 buffer in K2HPO4 form = (Moles of K2HPO4 / Moles of PO4²-) × 100
% of PO4 buffer in K2HPO4 form = (0.0015 mol / 0.0027 mol) × 100
% of PO4 buffer in K2HPO4 form ≈ 55.56%
Therefore, approximately 55.56% of the PO4 buffer is in the K2HPO4 form.
To find the percentage of the PO4 buffer in the K2HPO4 form, we can use the Henderson-Hasselbalch equation for buffer solutions:
pH = pKa + log([salt]/[acid])
In this case, the acid refers to KH2PO4, and the salt refers to K2HPO4.
First, let's determine the concentrations of the acid and salt. We know that the concentration of both KH2PO4 and K2HPO4 is 0.05 M. Since the volumes are different, we need to calculate the moles of each compound separately.
For KH2PO4:
moles of KH2PO4 = concentration (M) × volume (L)
= 0.05 M × 0.024 L
= 0.0012 mol
For K2HPO4:
moles of K2HPO4 = concentration (M) × volume (L)
= 0.05 M × 0.03 L
= 0.0015 mol
Next, we need to calculate the ratio of the salt to the acid.
ratio = (moles of K2HPO4) / (moles of KH2PO4)
= 0.0015 mol / 0.0012 mol
= 1.25
Now, we can substitute the ratio into the Henderson-Hasselbalch equation:
pH = pKa + log(ratio)
The pKa for the phosphate buffer system is approximately 7.2. Therefore:
pH = 7.2 + log(1.25)
Now, we need to solve for the pH.
To do that, we can rearrange the equation:
log(1.25) = pH - 7.2
Taking the antilog of both sides:
1.25 = 10^(pH - 7.2)
Now, let's solve for pH:
pH - 7.2 = log(1.25)
pH = log(1.25) + 7.2
pH ≈ 7.2 + 0.097
pH ≈ 7.297
Therefore, the buffer solution will have a pH of approximately 7.297 when mixed.
To determine the percentage of the PO4 buffer in the K2HPO4 form, we need to calculate the concentration of K2HPO4/(K2HPO4 + KH2PO4) and multiply by 100.
concentration of K2HPO4/(K2HPO4 + KH2PO4) = (0.0015 mol / (0.0012 mol + 0.0015 mol)) × 100
= (0.0015 mol / 0.0027 mol) × 100
= 55.55%
Therefore, approximately 55.55% of the PO4 buffer is in the K2HPO4 form.