A buffer solution of acetic acid and sodium acetate reacts with hydrochloric acid (or sodium hydroxide) according to the reactions below. Calculate the pH of 70.0mL of a buffer solution formed from 3.0 M sodium acetate and 4.0 M acetic acid after the addition. Don't forget - use moles instead of Molarity Ⓒ HCI+ NaC₂H302→ NaCl +HC₂H3O₂ NaOH + HC₂H302 → H₂O + NaC2H3O2 Moles HCI or NaOH PH 10.0 mL of 3.00 M HCI 30.0 mL of 3.00 M HCI 60.0 mL of 3.00 M HCI 10.0 mL of 3.00 M NaOH 30.0 mL of 3.00 M NaOH 60.0 mL of 3.00 M NaOH pH = pka + log pH =pKa + log salt + base acid-base) salt-acid- acid + acid,
A buffer solution of acetic acid and sodium acetate reacts with hydrochloric acid (or sodium hydroxide) according to the reactions below. Calculate the pH of 70.0mL of a buffer solution formed from 3.0 M sodium acetate and 4.0 M acetic acid after the addition. Don't forget - use moles instead of Molarity Ⓒ HCI+ NaC₂H302→ NaCl +HC₂H3O₂ NaOH + HC₂H302 → H₂O + NaC2H3O2 Moles HCI or NaOH PH 10.0 mL of 3.00 M HCI 30.0 mL of 3.00 M HCI 60.0 mL of 3.00 M HCI 10.0 mL of 3.00 M NaOH 30.0 mL of 3.00 M NaOH 60.0 mL of 3.00 M NaOH pH = pka + log pH =pKa + log salt + base acid-base) salt-acid- acid + acid,
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:4) A buffer solution of acetic acid and sodium acetate reacts with hydrochloric acid (or
sodium hydroxide) according to the reactions below. Calculate the pH of 70.0mL of a
buffer solution formed from 3.0 M sodium acetate and 4.0 M acetic acid after the
addition. Don't forget - use moles instead of Molarity
HCI+ NaC₂H302→ NaCl +HC2H302
NaOH + HC₂H302 → H₂O + NaC₂H3O2
Moles HC1 or NaOH PH
10.0 mL of 3.00 M HCI
30.0 mL of 3.00 M HCI
60.0 mL of 3.00 M HCI
10.0 mL of 3.00 M NaOH
30.0 mL of 3.00 M NaOH
60.0 mL of 3.00 M NaOH
(salt + base
pH = pka + log acid-base)
pH = pka + log
salt - acid
acid + acid/
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