Answer all questions (1-4) Answer all questions (1-4) Zoom in to veiw clearer. Q

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Answer all questions (1-4) Answer all questions (1-4) Zoom in to veiw clearer. QUESTION 1 You prepare a phosphate buffer by taking 5.5 mL of 0.75 M NaH2PO4 and 4.5 ml of 0.75 M Na2HPO4 and dilute to a final volume of 200. mL using distilled water. What is the concentration of NaH2P04 in the buffer solution? QUESTION 2 You prepare a phosphate buffer by taking 3.5 ml of 0.75M NaH2PO4 and 6.5 mL of 0.75 M Na2HP04 and dilute to a final volume of 200. mL using distilled water. What is the concentration of Na2HPO4 in the buffer solution? QUESTION 3 The pH of a buffer solution containing 0.150 M Na2HPO4 and 0.430 M NaH2PO4 is 6.75. Using the Henderson-Hasselbach Equation, calculate the pKa. QUESTION 4 The best buffer will Have the smallest change in pH per drop added o cannot be determined from change in pH per drop added Have a pH of 7.00 Have the greatest change in pH per drop added Click Save and Submit to sqve and submit. Click Save All Anscers to save all answers Save All Answers

Explanation / Answer

Ans. #1. Using           C1V1 (original solution) = C2V2 (final solution)

            Or, 0.75 M x 5.5 mL = C2 x 200.0 mL

            Or, C2 = (0.75 M x 5.5 mL) / 200.0 mL

            Hence, C2 = 0.020625 M

Therefore, [NaH2PO4] in final diluted solution = 0.020625 M

#2. Using       C1V1 (original solution) = C2V2 (final solution)

            Or, 0.75 M x 6.5 mL = C2 x 200.0 mL

            Or, C2 = (0.75 M x 6.5 mL) / 200.0 mL

            Hence, C2 = 0.024375M

Therefore, [Na2HPO4] in final diluted solution = 0.024375 M

#3. Na2HPO4 ------> HPO42-(aq) + 2Na+(aq)                  - monoprotic, Conjugate Base

NaH2PO4 ------> H2PO4-(aq) + Na+(aq)               - diprotic form, weak acid

The diprotic phosphate H2PO­2- acts as weak acid (pKa = 7.21). It can donate a proton to become HPO42- as follow-

            H2PO4-(aq) <---------> HPO42-(aq) + H+

Now, using Henderson- Hasselbalch equation

            pH = pKa + log ([A-] / [AH])                    - equation 1

                        where, [A-] = [Na2HPO4]                 

[AH] = [NaH2PO4]

Now,

            Putting the values in equation 1-

            6.75 = pKa + log ([Na2HPO4] / [NaH2PO4])

            Or, 6.75 = pKa + log (0.150 / 0.430) = pKa + (-0.46)

            Or, pKa = 6.75 + 0.46

            Hence, pKa = 7.21

#4. Correct option- a. Have the smallest change in pH per drop added.

A buffer is characterized by its ability to resist/minimize change in pH when a small quantity of acid or base is added to it. Smaller is the change in pH upon addition of acid/base, better is the buffer.

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