Calculate the masses of the conjugate acid and base needed to prepare each of th

Question

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. (pKa of H2PO4

– Æ 6.64) Assume the availability of the phosphate salts

listed in the table on the right.

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. (pKa of H2P 6.64) 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 KH2PO K2HPO4 NaH2PO.H20 Na,HPO-7H2o Molar mass (gmol-1) 136.09 174.2 137.99 268.09 KH2PO4/K2HPO4 (0.25 M, 100 mL) NaH2PO4/Na2HPO4 (0.25 M, 100 mL) pH 6.50 (Solution 1) pH 6.50 (Solution ) pH 6.50 (Solution 4) pH 6.70 (Solution 2) pH 6.70 (Solution 5) pH 6.80 (Solution 3) pH 6.80 (Solution 6) pH 6.50 pH 6.70 pH 6.80 Moles of H2PO4 Moles of HPO2 Solution 1 Solution 2 Solution 3 Grams of KH2PO Grams of K2HPO4 Solution 4 Solution 5 Solution 6 Grams of NaH,P H,o Grams of Na2HPO4-7H20 Suppose you added 2.00 mL of 0.250 M HCl to the pH 6.80 hydrogen phosphate buffer that was made in Question 1. What is the pH of the resulting buffer solution? Show all work. 2.

Explanation / Answer

1: Here the buffer action is due to the following equilibrium reaction

H2PO4-(aq) + H2O(l) < ---- > HPO42-(aq) + H3O+, pKa = 6.64

Here the acid is H2PO4-(aq) and the salt is HPO42-(aq).

We can find the moles of  H2PO4-(aq) and HPO42-(aq) required by applying Hendersen equation.

For pH = 6.50:

Here the required pH = 6.50

Applying Hendersen equation

pH = pKa + log [salt] / [acid] = pKa + log [HPO42-(aq)] / [H2PO4-(aq)]

=> 6.50 = 6.64 +  log [HPO42-] / [H2PO4-] ----- (1)

=>    log [HPO42-] / [H2PO4-]  = 6.50 - 6.64 = - 0.14

=>   [HPO42-] / [H2PO4-] = antilog (- 0.14) = 10-0.14 = 0.7244 ---- (2)

Here we need to prepare 100 mL of 0.25M HPO42- / H2PO4- buffer solution, V = 100 mL = 0.100 L

=> [(moles of HPO42-) / 100 mL] / [(moles of H2PO4-) / 100 mL] = 0.7244

=> (moles of HPO42-) / (moles of H2PO4-) = 0.7244

=>  (moles of HPO42-) = 0.7244 x(moles ofH2PO4-) ---------- (3)

Since we need to Prepare 100 mL of 0.25M HPO42- / H2PO4- buffer solution,

Total moles of HPO42- and H2PO4- = MxV = 0.25M x 0.100 L = 0.025 mol

=> (moles of HPO42-) + (moles of H2PO4-) = 0.025 mol ------(4)

By solving eqn (3) and (4) we get

Moles of HPO42- = 0.0105 mol (answer)

Moles of  H2PO4- = 0.0145 mol (answer)

1 mole of KH2PO4 contains 1 mole of  H2PO4-

Hence mass of KH2PO4 = moles of H2PO4- x Molecular mass of KH2PO4

= 0.0145 mol x (136.09 g/mol) = 1.973 g KH2PO4 (answer)

1 mole of K2HPO4 also contains 1 mole of  HPO42-

Hence mass of K2HPO4  = moles of  HPO42- x Molecular mass of K2HPO4

= 0.0105 mol x (174.2 g/mol) = 1.829 g K2HPO4 (answer)

1 mole of NaH2PO4.H2O contains 1 mole of  H2PO4-

Hence mass of NaH2PO4.H2O = moles of H2PO4- x Molecular mass of NaH2PO4.H2O

= 0.0145 mol x (137.99 g/mol) = 2.001 g NaH2PO4.H2O (answer)

Similarly mass of Na2HPO4.7H2O = moles of  HPO42- x Molecular mass of Na2HPO4.7H2O

= 0.0105 mol x 268.09 g/mol = 2.815 g  Na2HPO4.7H2O (answer)

For pH = 6.70:

Applying Hendersen equation

[HPO42-] / [H2PO4-] = antilog (0.06) = 100.06 = 1.148

(moles of HPO42-) + (moles of H2PO4-) = 0.025 mol

By solving the above two equations

Moles of HPO42- = 0.0134 mol (answer)

Moles of  H2PO4- = 0.0116 mol (answer)

Now we can calculate all the masses in a similar way we calculated for the pH of 6.50.

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