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Table 4 Volume of Moles of Moles of moles of sodium acètate Log pK, pH Acetic Acid Acetic Sodium moles acetic acid mol Na Ac of Acid (mol) Acetate (mol) Solution HAC mol HAC Added (mL) 2.00 12.5 mol 0.001 mol 0.00008 4.74 4.00 6.25 mol 0.001 mol 0.00016 4.74 6.00 4.17 mol 0.001 mol 0.00024 4.74 8.00 3.13 mol 0.001 mol 0.00032 4.74

Table 4 Volume of Moles of Moles of moles of sodium acètate Log pK, pH Acetic Acid Acetic Sodium moles acetic acid mol Na Ac of Acid (mol) Acetate (mol) Solution HAC mol HAC Added (mL) 2.00 12.5 mol 0.001 mol 0.00008 4.74 4.00 6.25 mol 0.001 mol 0.00016 4.74 6.00 4.17 mol 0.001 mol 0.00024 4.74 8.00 3.13 mol 0.001 mol 0.00032 4.74

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter13: The Chemistry Of Solutes And Solutions
Section: Chapter Questions
Problem 132QRT: A 0.109 mol/kg aqueous solution of formic acid, HCOOH, freezes at −0.210 °C. Calculate the...
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I need help finding the log and pH.

Table 4 Volume of Moles of Moles of moles of sodium acètate Log pK, pH Acetic Acid Acetic Sodium moles acetic acid mol Na Ac of Acid (mol) Acetate (mol) Solution HAC mol HAC Added (mL) 2.00 12.5 mol 0.001 mol 0.00008 4.74 4.00 6.25 mol 0.001 mol 0.00016 4.74 6.00 4.17 mol 0.001 mol 0.00024 4.74 8.00 3.13 mol 0.001 mol 0.00032 4.74
Transcribed Image Text:Table 4 Volume of Moles of Moles of moles of sodium acètate Log pK, pH Acetic Acid Acetic Sodium moles acetic acid mol Na Ac of Acid (mol) Acetate (mol) Solution HAC mol HAC Added (mL) 2.00 12.5 mol 0.001 mol 0.00008 4.74 4.00 6.25 mol 0.001 mol 0.00016 4.74 6.00 4.17 mol 0.001 mol 0.00024 4.74 8.00 3.13 mol 0.001 mol 0.00032 4.74
Expert Solution
Step 1

Calculate the log(mol NaAc/mol HAc) value,

 

            log(mol NaAcmol HAc) = log (0.00008) = -4.1log(mol NaAcmol HAc) =  log (0.00016) = -3.8 log(mol NaAcmol HAc) = log (0.00024) = -3.6log(mol NaAcmol HAc) =  log (0.00032) = -3.5

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