Ka of acetic acid equals 1.8*10^-5 The buffer solution is 50 mL of 0.2 M CH3CO2H and 50 mL of 0.2 M NaCH3CO2 together in a beaker. What is the theoretical pH of each soluton? a. 50 mL of boiled deionized water + 5 mL of 0.2 M HCl b. 50 mL of boiled deionized water + 5 mL of 0.2 M NaOH c. 50 mL of buffer + 5 mL of 0.2 M HCl d. 50 mL of buffer + 5 mL of 0.2 M NaOH
Ka of acetic acid equals 1.8*10^-5 The buffer solution is 50 mL of 0.2 M CH3CO2H and 50 mL of 0.2 M NaCH3CO2 together in a beaker. What is the theoretical pH of each soluton? a. 50 mL of boiled deionized water + 5 mL of 0.2 M HCl b. 50 mL of boiled deionized water + 5 mL of 0.2 M NaOH c. 50 mL of buffer + 5 mL of 0.2 M HCl d. 50 mL of buffer + 5 mL of 0.2 M NaOH
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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.141QP: Ka for formic acid is 1.7 104 at 25C. A buffer is made by mixing 529 mL of 0.465 M formic acid,...
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Ka of acetic acid equals 1.8*10^-5
The buffer solution is 50 mL of 0.2 M CH3CO2H and 50 mL of 0.2 M NaCH3CO2
together in a beaker.
What is the theoretical pH of each soluton?
a. 50 mL of boiled deionized water + 5 mL of 0.2 M HCl
b. 50 mL of boiled deionized water + 5 mL of 0.2 M NaOH
c. 50 mL of buffer + 5 mL of 0.2 M HCl
d. 50 mL of buffer + 5 mL of 0.2 M NaOH
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Step 1: Henderson's equation
VIEWStep 2: pH of 50 mL of boiled deionised water + 5 mL of 0.2 M HCl
VIEWStep 3: pH of 50 mL of boiled deionized water + 5 mL of 0.2 M NaOH
VIEWStep 4: pH of 50 mL of buffer (CH3COOH and CH3COO-) + 5 mL of 0.2 M HCl
VIEWStep 5: pH of 50 mL of buffer (CH3COOH and CH3COO-) + 5 mL of 0.2 M NaOH
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