Answer:
pH1=4.98
pH2=5.06
pH3=4.90
Explanation:
Hello there!
In this case, it is possible to analyze the problem according to the Henderson-Hasselbach equation:
[tex]pH=pKa+log(\frac{[base]}{[acid]} )[/tex]
Whereas the pKa = 4.88 and the base is identified as the sodium propionate and the acid as the propionic acid. Now, for the pH of the buffer we simply plug the given data in the formula to obtain:
[tex]pH=4.88+log(\frac{0.25mol/1.50dm^3}{0.20mol/1.50dm^3} )\\\\pH=4.98[/tex]
Next, since the addition of 0.02 mol NaOH results in the consumption of 0.02 moles of acid and the formation of 0.02 moles of base, we use the following modification for the HH equation:
[tex]pH=4.88+log(\frac{(0.25+0.02)mol/1.50dm^3}{(0.20-0.02)mol/1.50dm^3} )\\\\pH=5.06[/tex]
Which of course, increases the pH as NaOH is a strong base. Finally, for the addition of 0.02 mol HI, a consumption of the base and formation of the acid are implied since HI is a strong acid, for which the pH turns out to be less than the initial 4.98:
[tex]pH=4.88+log(\frac{(0.25-0.02)mol/1.50dm^3}{(0.20+0.02)mol/1.50dm^3} )\\\\pH=4.90[/tex]
Best regards!
