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Question

??76960? Cengag x ?Desmos x W List of b x( G How Ma X' G A Buffer x 7 Table of PKa eAssignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSessionLocator-as; Window Help Use the References to access important values if needed for this question. A buffer solution contains 0.417 M acetic acid and 0.216 M sodium acetate. Ir0.0342 moles of sodium hydroxide are added to 150 ml of this buffer, what is the pl of the resulting solution? (Assume that the volume does not change upon adding sedium hydroxide pli- Retry Entire Group 7 more group attempts remalning Submit Answer

Explanation / Answer

Answer – We are given, [CH3COOH] = 0.417 M

[NaCH3COO] = 0.216 M, volume = 150 mL , moles of NaOH = 0.0342 moles

Now first we need to calculate moles of acid and its conjugate base-

Moles of CH3COOH = 0.417 M X 0.150 L

                                  = 0.0625 moles

Moles of NaCH3COO = 0.216 M X 0.150 L

                                    = 0.0324 moles

when we added NaOH the moles of acid decrease and its conjugate base increase

Moles of CH3COOH = 0.0625 moles – 0.0342 mole = 0.0284 moles

moles of CH3COO- = 0.0324 moles + 0.0342 moles = 0.0666 moles

Total volume = 150 mL

so, [CH3COOH] = 0.0284 moles / 0.150 L = 0.189 M

[CH3COO-] = 0.0666 mol / 0.150 L = 0.444 M

Now we need to use Henderson Hasselbalch equation

pH = pKa + log [CH3COO-] / [CH3COOH]

      = 4.75 + log 0.444 / 0.189

      = 5.12

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