Exercise 16.49 Part A For 510 0 mL of pure water. calculate the initial pH and t
ID: 905246 • Letter: E
Question
Exercise 16.49 Part A For 510 0 mL of pure water. calculate the initial pH and the final pH after adding 0.020 mol of Ha Express your answers using two decimal places separated by a comma. PH initial , pH final = Part B For 510.0 mL of a buffer solution that is 0.120 M in HC2H3O2 and 0.100 M in NaC2H3O2. calculate the initial pH and the final pH after adding 0.020 mol of Ha Express your answers using two decimal places separated by a comma. PH initial , pH final = Part C For 510.0 mL of a buffer solution that is 0.160 Min CH3CH2NH2 and 0.140 Min CH3CH2NH3CL, calculate the initial pH and the final pH after adding 0.020 mol of Ha Express your answers using two decimal places separated by a comma.Explanation / Answer
Part A)
Initial pH = 7 (for pure water)
After adding HCl
Moles of HCl added = 0.02
Volume of water = 510 mL = 0.51 L
Conc. of HCl = 0.02 / 0.51 = 0.0392 M
=> [H+] = 0.0392 M
pH = - log [H+] = 1.41 = Final pH
Part B)
Ka for CH3COOH = 1.74 x 10^-5
pKa = 4.76
pH for a buffer = pKa + log (Salt / Acid)
=> pH = 4.76 + log (0.1 / 0.12) = 4.68 = pH before adding HCl
pH after adding HCl
pH = pKa + log (Salt - X / Acid + X)
X = Conc. of HCl added = 0.02 / 0.51 = 0.0392 M
=> pH = 4.76 + log (0.1 - 0.0392 / 0.12 + 0.0392) = 4.34 = pH after adding HCl
Part C)
pKb for CH3CH2NH2 = 3.3
pOH for a buffer = pKb + log (Salt / Base)
=> pOH = 3.3 + log (0.14 / 0.16) = 3.24
=> pH = 14 - 3.24 = 10.76 = pH before adding HCl
pH after adding HCl
pOH = pKb + log (Salt + X / Base - X)
X = Conc. of HCl added = 0.0392 M
pOH = 3.3 + log (0.14 + 0.0392 / 0.16 - 0.0392) = 3.47
=> pH = 14 - 3.47 = 10.53
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