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Please show all work for the following questions

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To make the buffer solution, 581 mL of 1.00 mol L-1 KH2PO4 is needed.

To calculate the amount of 1.00 mol L-1 KH2PO4 needed to make the 250 mL buffer solution, we need to use the Henderson-Hasselbalch equation.

The equation is pH = pKa + log([A-]/[HA]). Rearranging the equation, we have [A-]/[HA] = 10^(pH - pKa). In this case, the pKa is 7.21 and the desired pH is 6.87.

Plugging in these values and solving for [A-]/[HA], we get 10^(6.87 - 7.21) = 0.387. Since the ratio of [A-] to [HA] is 0.387, we can set up the following equation to calculate the volume of KH2PO4 solution needed:

(volume of KH2PO4 solution)/(volume of K2HPO4 solution) = (concentration of K2HPO4 solution)/(concentration of KH2PO4 solution) = 0.387

Since the volume of K2HPO4 solution is given as 1.50 L, we can solve for the volume of KH2PO4 solution:

volume of KH2PO4 solution = (0.387)(1.50 L) = 0.581 L

Converting to milliliters, the amount of 1.00 mol L-1 KH2PO4 needed to make the solution is 581 mL to three significant digits.

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