From your measurement above, you have found the pKa to be 10.3. \"Acid-free\" pa

Question

From your measurement above, you have found the pKa to be 10.3. "Acid-free" paper should have a pH between 8.7 and 10.2. What is the equilibrium concentration of carbonate in your buffer (in mM) if you add o 513 mM calcium carbonate to pulp that started with 0.359 mM acid (pH=3.37)? insert your numerical answer. From your measurement above, you have found the pKa to be 10.3. "Acid-free" paper should have a pH between 8.7 and 10.2. What is the pH of the buffer in your paper pulp if you add 0.643 mM calcium carbonate to pulp that started with 0.444 mM acid (pH=3.37)? insert your numerical answer.

Explanation / Answer

6) Initial acid concentration = 0.359 mM

added calcium carbonate = 0.513mM

[HA-] formed = 0.205 mM

[A^2-] formed = 0.154mM

This is a buffer solution

pH of buffer by Hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

     = 10.3 + log(0.154/0.205)

     =10.175

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