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.
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?
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