Question

Pre-Lab Assignment Hydrogen Phosphate Buffer System This pre-lab assignment is mandatory. You will receive a zero for the experiment if the pre-lab assignment is not completed before the beginning of the lab session. 1. Calculate the masses of the conjugate acid and base needed to prepare each of the six buffer solutions listed in the table on the left. (p of H,PO,= 6.70) Assume the availability of the phosphate salts listed in the table on the right. Fill out the worksheet below. Show all work on separate pages for full credit. Salt Molar mass (g mol-') KH,PO,/K HPO, (0.25 M, 100 mL.) pH 6.50 (Solution 1) pH 6.70 (Solution 2) pH 6.90 (Solution 3) NaH PO /Na HPO, (0.25 M, 100 ml.) pH 6.50 (Solution 4) pH 6.70 (Solution 5) pH 6.90 (Solution 6) KH,PO, K2HPO NaH,PO,H,O Na, HPO, -7H,0 136.09 174.2 137.99 268.09 pH 6.50 pH 6.70 pH 6.90 Moles of H,PO, Moles of HPO, Solution 1 Solution 2 Solution 3 Grams of KH,PO. Grams of K_HPO Solution 4 Solution 5 Solution 6 Grams of NaH,PO,H,O Grams of Na2HPO, 7H,0 2. Suppose you added 2.00 mL of 0.250 M HCl to the pH 6.70 hydrogen phosphate buffer that was made in Question 1. What is the pH of the resulting buffer solution Show all work

Answer 1

The first thing to calculate is the ratio of concentrations of the phosphate species. This can be calculated using the Henderson-Hasselbach equation:

$pH=pKa+log\left ( \frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]} \right )$

This can be re-arranged to calculate the ratio for the desired pHs:

$\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]}=10^{pH-pKa}$

$pH=6.50:\: \frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]}=10^{6.50-6.70}=0.631$

$pH=6.70:\: \frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]}=10^{6.70-6.70}=1$

$pH=6.90:\: \frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^{-}]}=10^{6.90-6.70}=1.58$

For each buffer, we have the same number of total moles, given by:

$n=M\cdot V(L)=0.25\frac{mol}{L}\cdot 0.1L=0.025\, moles$

From this value, we can calculate the concentration of each species. Bear in mind that, since both species are in the same volume of solution, the concentration ratio is the same as the moles ratio.

For the pH = 6.50 buffer:

$n_{HPO_{4}^{2-}}+n_{H_{2}PO_{4}^{-}}=0.025\, moles$

$0.631n_{H_{2}PO_{4}^{-}}+n_{H_{2}PO_{4}^{-}}=1.631n_{H_{2}PO_{4}^{-}}=0.025\, moles$

$n_{H_{2}PO_{4}^{-}}=\frac{0.025\, moles}{1.631}=0.0153\, moles$

$n_{HPO_{4}^{-}}=0.025\, moles-n_{H_{2}PO_{4}^{2-}}=0.0097\, moles$

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For the pH = 6.70 buffer:

$n_{HPO_{4}^{2-}}+n_{H_{2}PO_{4}^{-}}=0.025\, moles$

$n_{H_{2}PO_{4}^{-}}+n_{H_{2}PO_{4}^{-}}=2n_{H_{2}PO_{4}^{-}}=0.025\, moles$

$n_{H_{2}PO_{4}^{-}}=\frac{0.025\, moles}{2}=0.0125\, moles$

$n_{HPO_{4}^{-}}=0.025\, moles-n_{H_{2}PO_{4}^{2-}}=0.0125\, moles$

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For the pH = 6.90 buffer:

$n_{HPO_{4}^{2-}}+n_{H_{2}PO_{4}^{-}}=0.025\, moles$

$1.58n_{H_{2}PO_{4}^{-}}+n_{H_{2}PO_{4}^{-}}=2.58n_{H_{2}PO_{4}^{-}}=0.025\, moles$

$n_{H_{2}PO_{4}^{-}}=\frac{0.025\, moles}{2.58}=0.0097\, moles$

$n_{HPO_{4}^{-}}=0.025\, moles-n_{H_{2}PO_{4}^{2-}}=0.0153\, moles$

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With these values, we can calculate the mass of salts needed for each buffer. For solution 1:

KH₂PO₄: $m=n\cdot m_{molar}=0.0153\, moles\cdot 136.09\frac{g}{mol}=2.08\, g$

K₂HPO₄: $m=n\cdot m_{molar}=0.0097\, moles\cdot 174.2\frac{g}{mol}=1.69\, g$

Solution 2:

KH₂PO₄: $m=n\cdot m_{molar}=0.0125\, moles\cdot 136.09\frac{g}{mol}=1.70\, g$

K₂HPO₄: $m=n\cdot m_{molar}=0.0125\, moles\cdot 174.2\frac{g}{mol}=2.18\, g$

Solution 3:

KH₂PO₄: $m=n\cdot m_{molar}=0.0097\, moles\cdot 136.09\frac{g}{mol}=1.32\, g$

K₂HPO₄: $m=n\cdot m_{molar}=0.0153\, moles\cdot 174.2\frac{g}{mol}=2.67\, g$

Solution 4:

NaH₂PO₄: $m=n\cdot m_{molar}=0.0153\, moles\cdot 137.99\frac{g}{mol}=2.11\, g$

Na₂HPO₄: $m=n\cdot m_{molar}=0.0097\, moles\cdot 268.09\frac{g}{mol}=2.60\, g$

Solution 5:

NaH₂PO₄: $m=n\cdot m_{molar}=0.0125\, moles\cdot 137.99\frac{g}{mol}=1.72\, g$

Na₂HPO₄: $m=n\cdot m_{molar}=0.0125\, moles\cdot 268.09\frac{g}{mol}=3.35\, g$

Solution 6:

NaH₂PO₄: $m=n\cdot m_{molar}=0.0097\, moles\cdot 137.99\frac{g}{mol}=1.34\, g$

Na₂HPO₄: $m=n\cdot m_{molar}=0.0153\, moles\cdot 268.09\frac{g}{mol}=4.10\, g$

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If we added 2.00 mL of 0.250 M HCl, we would be adding:

$n=0.250\frac{mol}{L}\cdot 0.002L=0.0005\, moles$

This moles of acid would react:

$HPO_{4}^{2-}+H^{+}\rightarrow H_{2}PO_{4}^{-}$

So the number of moles of HPO₄^2- originally present would be decreased by 0.0005 moles and the number of H₂PO₄^- would be increased by that number, so we can recalculate the pH as (using the Henderson-Hasselbach equation):

$pH=6.70+log\left ( \frac{0.0125\, moles-0.0005\, moles}{0.0125\, moles+0.0005\, moles} \right )=6.67$