Calculate the pH of A) a solution that is 0.060 M in potassium propionate(C 2 H

Question

Calculate the pH of A) a solution that is 0.060 M in potassium propionate(C2H5COOK orKC3H5O2) and 0.085 M in propionicacid (C2H5COOH orHC3H5O2) B) A solution that is 0l075 in trimethylamine(CH3)3N and 0.10 M in trimethylammoniumchlroide (CH3)3NHCl C) A solution that is made by mixing 50.0 mL of 0.15 M aceticacid and 50.0 mL of 0.20 M sodium acetate Calculate the pH of A) a solution that is 0.060 M in potassium propionate(C2H5COOK orKC3H5O2) and 0.085 M in propionicacid (C2H5COOH orHC3H5O2) B) A solution that is 0l075 in trimethylamine(CH3)3N and 0.10 M in trimethylammoniumchlroide (CH3)3NHCl C) A solution that is made by mixing 50.0 mL of 0.15 M aceticacid and 50.0 mL of 0.20 M sodium acetate

Explanation / Answer

A) According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of propionic acid = 4.886      pH = pKa + log[base]/[acid]            = 4.2 + log[0.06M]/[0.085M]             =4.886 - 0.1512            = 4.7347            = 4.2 + log[0.06M]/[0.085M]             =4.886 - 0.1512            = 4.7347 B) According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of trimethylammonium ion = 4.2    pH = pKa + log[base]/[acid]            = 4.886 + log[0.075M]/[0.1M]             =4.886 - 0.1249            = 4.075 C) According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of acetic acid = 4.74    pH = pKa + log[base]/[acid]            = 4.74 + log[0.2M]/[0.15M]             =4.74 + 0.1249            = 4.8649 According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of trimethylammonium ion = 4.2    pH = pKa + log[base]/[acid]            = 4.886 + log[0.075M]/[0.1M]             =4.886 - 0.1249            = 4.075 C) According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of acetic acid = 4.74    pH = pKa + log[base]/[acid]            = 4.74 + log[0.2M]/[0.15M]             =4.74 + 0.1249            = 4.8649 According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of trimethylammonium ion = 4.2    pH = pKa + log[base]/[acid]            = 4.886 + log[0.075M]/[0.1M]             =4.886 - 0.1249            = 4.075 C) According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of acetic acid = 4.74    pH = pKa + log[base]/[acid]            = 4.74 + log[0.2M]/[0.15M]             =4.74 + 0.1249            = 4.8649 C) According to Henderson-Hasselbalch equation      pH = pKa + log[base]/[acid]    pKa of acetic acid = 4.74    pH = pKa + log[base]/[acid]            = 4.74 + log[0.2M]/[0.15M]             =4.74 + 0.1249            = 4.8649      pH = pKa + log[base]/[acid]    pKa of acetic acid = 4.74    pH = pKa + log[base]/[acid]            = 4.74 + log[0.2M]/[0.15M]             =4.74 + 0.1249            = 4.8649            = 4.74 + log[0.2M]/[0.15M]             =4.74 + 0.1249            = 4.8649

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