For each of the following solutions, calculate the initial pH and the final pH a
ID: 997512 • Letter: F
Question
For each of the following solutions, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH.
For 280.0 mL of pure water, calculate the initial pH and the finalpH after adding 0.0200 mol of NaOH.
Express your answers using two decimal places separated by a comma.
For 280.0 mL of a buffer solution that is 0.215 M in HCHO2and 0.300 M in KCHO2, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Ka=1.8104).
Express your answers using two decimal places separated by a comma.
For 280.0 mL of a buffer solution that is 0.2763 M in CH3CH2NH2 and 0.2563 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6104).
Express your answers using two decimal places separated by a comma.
Explanation / Answer
A)
initial pH of pure water in which [H+]= [OH-] = 7.00
after adding 0.0200 mol NaOH
[OH-]= 0.0200 mol/ 0.280 L=0.0714 M
pOH = - log 0.0714 = 1.14
pH = 14 - pOH = 14 - 1.14 = 12.85
B)
pKa of formic acid = 3.74 ( check this value on your book)
pH = pKa + log [HCOO-]/ [HCOOH]
pH = 3.74 + log 0.300/ 0.215 =3.884 ( initial pH)
moles HCOOH = 0.280 L x 0.215 M=0.0602
moles HCOO- = 0.280 L x 0.300 M=0.0840
after adding 0.005 mol OH- the moles of HCOOH will decreases by 0.0200 and the moles of HCOO- will increase by 0.0200 by the reaction
HCOOH + OH- = HCOO- + H2O
moles HCOOH = 0.0574 - 0.0200 =0.0374
moles HCOO- = 0.0840 + 0.0200 =0.104
concentration HCOOH = 0.0374 / 0.280 L = 0.1335 M
concentration HCOO- = 0.104 / 0.280 L=0.371 M
pH = 3.74 + log 0.371 / 0.133=4.18 ( final pH)
c)
pKb of ethylamine = 3.37 ( check this value on your book)
pOH = 3.37 + log 0.256/ 0.276= 3.33
pH = 14 - pOH =10.66 ( initial pH)
moles ethylamine = 0.280 L x 0.276 M=0.0772
moles ethylammonium = 0.280 L x 0.256 M=0.0716
CH3CH2NH3++ OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.0716 - 0.0200 =0.0516
moles CH3CH2NH2 = 0.0772 + 0.0200 =0.0972
concentration CH3CH2NH3+ = 0.0516/ 0.280=0.0.184 M
concentration CH3CH2NH2 = 0.0972/0.280 L=0.347 M
pOH = 3.37 + log 0.184 / 0.347 = 3.09
pH = 14 - 3.09=10.90
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