Calculate the change in pi I that occurs when 1.20 mmol of a strong acid is adde
ID: 1016311 • Letter: C
Question
Calculate the change in pi I that occurs when 1.20 mmol of a strong acid is added to 100 mL of the solutions listed below. (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) H_2 O Change in pH = 0.0700 M HCl. Change in pH = 0.0700 M NaOH. Change in pH = 0.0700 M CH_3CH_2CH_2COOH. Change in pH = 0.0700 M CH_3 CH_2 CH_2 COONa. Change in pH 0.0700 M CH_3CH_2CH_2COOH + 0.0700 M CH_3CH_2CH_2 COONa. Change in pH = 0.700 M CH_3CH_2CH_2COOH + 0.700 M CH_3CH_2CH_2COONa. Change in pH =Explanation / Answer
a)
original [H+] = 10^-7 M
original pH = 7
added [H+] = 1.2*10^-3 mol / 0.1 L = 0.012 M
final [H+] = 0.012 M (roughly)
final pH = -log [H+]
= -log (0.012)
= 1.92
change in pH= 7 - 1.92 = 5.08
b)
original [H+] = 0.07 M
original pH= -log (0.07) = 1.15
added [H+] = 1.2*10^-3 mol / 0.1 L = 0.012 M
Final [H+] = 0.012 + 0.07 M = 0.082 M
final pH = -log [H+]
= -log (0.082)
= 1.086
change in pH = 1.15 - 1.086 = 0.064
c)
original [OH-] = 0.07 M
original pOH= -log (0.07) = 1.15
original pH = 14 - 1.15 = 12.85
added [H+] = 1.2*10^-3 mol / 0.1 L = 0.012
H+ and OH- will neutralise each other
Final [OH-] = 0.07 M - 0.012= 0.058 M
final pOH = -log [OH-]
= -log (0.058)
= 1.237
final pH = 14 - 1.237 = 12.763
change in pH = 12.85 - 12.763 = 0.087
Rest all are weak acids . You need to provide the dissociation constantr for the same
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