1 A buffer is prepared by mixing 50.0 mL of 0.100 M acetic acid with 30.0 mL of

Question

1 A buffer is prepared by mixing 50.0 mL of 0.100 M acetic acid with 30.0 mL of 0.100 M sodium hydroxide and 20.0 mL of distilled water. Estimate the concentration of acetic acid in the resulting buffer. Report your answer in moles/liter, but do not include units with your answer. 2 Determine the concentration of acetate ions in the buffer solution. Report your answer in moles per liter, but do not include units in your answer. 3 Estimate the pH of the buffer solution using the Henderson-Hasselbalch equation. 4 Using the Henderson-Hasselbalch equation, estimate the pH of the buffer if an additional 10.0 mL of 0.100 M sodium hydroxide is added. 5 Estimate the pH of the original buffer if 10.0 mL of 0.100 M hydrochloric acid is added.

Explanation / Answer

                  CH3COOH+ NaOH---> CH3COONa+ H2O

moles of acetic acid = 0.1*50/1000=0.005 moles moles of NaOH= 0.1*30/1000=0.003 moles

excess is acetic acid by = 0.005-0.003 =0.002 moles

Volume after mixing acetic acid and sodium hydroxide= 50+30= 80ml =80/1000 L=0.08L

Concentration of acetic acid = 0.002/0.08= 0.025

2. Moles of sodium acetate formred= 0.003 ( same as moles of Sodium hydroxide)

Volume after mixing Acettic acid and NaOH= 0.08L, concentration of Sodium acetate= 0.003/0.08=0.0375

3) pH= Pka+ log[A-]/[HA] before addition of additional 10ml of 0.1M NaOH, A- sodium acetate ion and HA= acetic acid

Ka of acetic acid = 1.75*10-5, pKa=-logKa= 4.76+log[0.0375/0.025)=4.93

b) when 10ml of 0.1M NaOH is added, it adds to formation of additional Sodium acetate and decrease in acetic acid

total moles of sodium hydroxide= 0.1*30/1000+0.1*10/1000= 0.004 moles.

moles of acetic acid remaining after reaction with NaOH ( CH3COOH+ NaOH---> CH3COONa+H2O)

moles of CH3COOH remaining= 0.005-0.004= 0.001 moles   volume after mixing 10ml of NaOH= 50+30+10=90ml =0.09L concentration of CH3COOH= 0.001/0.09=0.011

Concentration of acetate ion (A-)= 0.004/0.09=0.044

pH= pKa+log[A-]/[HA] = 4.93+log(0.044/0.011)=5.532

when HCl is added, this reacts with NaOH as HCl+ NaOH---> NaCl+ H2O

moles of HCl= 0.1*10/1000=0.001 moles of NaOH= 0.003 moles

Moles of NaOH reamining after the reaction with HCl= 0.003-0.001= 0.002 moles

Addition of HCl reduced NaOH used for neutralization of acetic acid

hence moles of sodiuum acetate formed now =0.002moles

Volume of mixing = 90ml, concentration of A - = 0.002/0.09

acetic acid remaining after forming sodium acetate on reaction with NaOH= 0.1*50/1000-0.002=0.003

concentration of acetic acid =0.003/0.09

pH= 4.93+log(0.002/0.003)= 4.753

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