DATA ANALYSIS C.1 Did the acidic and basic buffer solutions maintain a relativel

Question

DATA ANALYSIS C.1 Did the acidic and basic buffer solutions maintain a relatively constant pH after the addition of HC1 (0.5 M) and NaOH (0.5 M)? Describe vour observations Attach to this report two excel plots showing the evolution of the measured pHH data for the two buffers versus the cumulative volume of HCl or NaOH added Note: Each plot must contain the data from the two buffers C.2 How do your experimental data agree with the expected working range of each buffer that can be estimated from the pKa value? Explain C.3 Observe the trends in pH for both buffer solutions, outside their buffering range, ie upon adding a considerable amount of a strong base or acid. - After adding 10 mL of NaOH (0.5 M), you notice that the pH values of both acidic and basic buffer solutions equally converge to the value that can be predicted from the effective concentration of the strong base (NaOH) in the final mixture. - However, when 10 mL HC1(0.5 M) is added to each buffer, only the pH value for the acetic acid/acetate system converges to that predicted by the effective concentration of the strong acid (HCI) in the final mixture. Why does the pH of the bicarbonatel/carbonate system drop only to a mild value, close to neutral? Explain this observation.

Explanation / Answer

Yes, the addition of 0.5 M HCl or 0.5 M NaOH to the acidic and basic buffer solutions maintains the relatively constant pH.

Example: (i) There is 1 mmol of sodium acetate and 0.7 mmol of acetic acid in the acidic buffer solution.

According to Henderson-Hasselbulch equation: pH = pKa + Log(nsodium acetate/nacetic acid)

i.e. pH = 4.74 + Log(1/0.7)

i.e. pH = 4.895

If you add 1 mL of 0.5 M HCl to it, i.e. nHCl = 1 mL * 0.5 mmol/mL = 0.5 mmol

Now, pH = 4.74 + Log{(1-0.5)/(0.7+0.5)}

i.e. pH = 4.360

(ii) There is 1 mmol of ammonium chloride and 0.7 mmol of ammonium hydroxide in the basic buffer solution.

According to Henderson-Hasselbulch equation: pH = pKa + Log(nammonium hydroxide/nammonium chloride)

i.e. pH = 9.26 + Log(0.7/1)

i.e. pH = 9.105

If you add 1 mL of 0.5 M NaOH to it, i.e. nNaOH = 1 mL * 0.5 mmol/mL = 0.5 mmol

Now, pH = 9.26 + Log{(0.7+0.5)/(1-0.5)}

i.e. pH = 9.640

Hence, the pH of the solution maintains the relatively constant after the addition of HCl (0.5 M) or NaOH (0.5 M).

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