1) A buffer solution contains 0.329 M NH 4 Cl and 0.482 M NH 3 ( ammonia ). Dete
ID: 705180 • Letter: 1
Question
1) A buffer solution contains 0.329 M NH4Cl and 0.482 M NH3 (ammonia). Determine the pH change when 0.130 mol HNO3 is added to 1.00 L of the buffer.
pH after addition ? pH before addition = pH change = ???
2) A buffer solution contains 0.363 M C6H5NH3Cl and 0.212 M C6H5NH2 (aniline). Determine the pH change when 0.050 mol KOH is added to 1.00 L of the buffer.
pH after addition ? pH before addition = pH change = ????
3) A buffer solution contains 0.360 M NaHCO3 and 0.397 M K2CO3. Determine the pH change when 0.108 mol HNO3 is added to 1.00 L of the buffer.
pH change = ???
Explanation / Answer
1) Consider the ionization of ammonium chloride.
NH4Cl (aq) -------> NH4+ (aq) + Cl- (aq)
Cl- is a spectator ion and the buffer system consist of NH4+ and NH3. The ionization of NH4+ is given below.
NH4+ (aq) --------> NH3 (aq) + H+ (aq); pKa = 9.25
The pH of the buffer is given by the Henderson-Hasslebach equation as
pH = pKa + log [NH3]/[NH4+]
= 9.25 + log (0.482 M)/(0.329 M)
= 9.25 + log (1.4650)
= 9.25 + (0.1658)
= 9.4158 ? 9.41 (ans).
We have 1.00 L of the buffer solution; therefore,
Moles NH4+ = (volume of buffer in L)*(molarity of NH4+) = (1.00 L)*(0.329 M) = 0.329 mole
Moles NH3 = (volume of buffer in L)*(molarity of NH3) = (1.00 L)*(0.482 M) = 0.482 mole.
HNO3 ionizes completely to furnish H+ which reacts with NH3 as below.
HNO3 (aq) --------> H+ (aq) + NO3- (aq)
H+ (aq) + NH3 (aq) --------> NH4+ (aq)
As per the stoichiometric equation,
moles H+ = moles HNO3 = moles NH3 neutralized = moles NH4+ formed = 0.130 mole.
Moles NH3 after the addition of HNO3 = (0.482 – 0.130) mole = 0.352 mole.
Moles NH4+ formed after the addition = (0.329 + 0.130) mole = 0.459 mole.
Since the volume of the buffer is 1.00 L, therefore,
[NH3] = (moles of NH3)/(volume of the buffer) = (0.352 mole)/(1.00 L) = 0.352 M.
[NH4+] = (moles of NH4+)/(volume of the buffer) = (0.459 mole)/(1.00 L) = 0.459 M.
Use the Henderson-Hasslebach equation to find the pH.
pH = pKa + log [NH3]/[NH4+]
= 9.25 + log (0.352 M)/(0.459 M)
= 9.25 + log (0.7669)
= 9.25 + (-0.1153)
= 9.25 – 0.1153 = 9.1347 ? 9.13 (ans).
Change in pH of the buffer after the addition of HNO3 = (new pH) – (old pH) = (9.13) – (9.41) = -0.28 (ans).
2) Consider the ionization of anilium chloride.
C6H5NH3+Cl- (aq) -------> C6H5NH3+ (aq) + Cl- (aq)
Cl- is a spectator ion and the buffer system consist of C6H5NH3+ and C6H5NH2. The ionization of C6H5NH3+ is given below.
C6H5NH3+ (aq) --------> C6H5NH2 (aq) + H+ (aq); pKa = 4.60
The pH of the buffer is given by the Henderson-Hasslebach equation as
pH = pKa + log [C6H5NH2]/[C6H5NH3+]
= 4.60 + log (0.212 M)/(0.363 M)
= 4.60 + log (0.5840)
= 4.60 + (-0.2336)
= 4.60 – 0.2336 = 4.3664 ? 4.37 (ans).
We have 1.00 L of the buffer solution; therefore,
Moles C6H5NH3+ = (volume of buffer in L)*(molarity of C6H5NH3+) = (1.00 L)*(0.363 M) = 0.363 mole
Moles C6H5NH2 = (volume of buffer in L)*(molarity of C6H5NH2) = (1.00 L)*(0.212 M) = 0.212 mole.
KOH ionizes completely to furnish OH- which reacts with C6H5NH3+ as below.
KOH (aq) --------> K+ (aq) + OH- (aq)
C6H5NH3+ (aq) + OH- (aq) ---------> C6H5NH2 (aq) + H2O (l)
As per the stoichiometric equation,
moles OH- = moles KOH = moles C6H5NH3+ neutralized = moles C6H5NH2 formed = 0.050 mole.
Moles C6H5NH3+ after the addition of KOH = (0.363 – 0.050) mole = 0.313 mole.
Moles C6H5NH2 formed after the addition = (0.212 + 0.050) mole = 0.262 mole.
Since the volume of the buffer is 1.00 L, therefore,
[C6H5NH2] = (moles of C6H5NH2)/(volume of the buffer) = (0.262 mole)/(1.00 L) = 0.262 M.
[C6H5NH3+] = (moles of C6H5NH3+)/(volume of the buffer) = (0.313 mole)/(1.00 L) = 0.313 M.
Use the Henderson-Hasslebach equation to find the pH.
pH = pKa + log [C6H5NH2]/[C6H5NH3+]
= 4.60 + log (0.262 M)/(0.313 M)
= 4.60 + log (0.8371)
= 4.60 + (-0.0772)
= 4.60 – 0.0722 = 4.5228 ? 4.52 (ans).
Change in pH of the buffer after the addition of KOH = (new pH) – (old pH) = (4.52) – (4.37) = 0.15 (ans).
3) Consider the ionizations of NaHCO3 and K2CO3
NaHCO3 (aq) -------> Na+ (aq) + HCO3- (aq)
K2CO3 (aq) --------> 2 K+ (aq) + CO32- (aq)
Na+ and K+ are spectator ions and the buffer system consists of HCO3- and CO32-. The ionization of HCO3- is given below.
HCO3- (aq) --------> CO32- (aq) + H+ (aq); pKa = 10.32
The pH of the buffer is given by the Henderson-Hasslebach equation as
pH = pKa + log [CO32-]/[HCO3-]
= 10.32+ log (0.397 M)/(0.360 M)
= 10.32 + log (1.1028)
= 10.32 + (0.0425)
= 10.3625 ? 10.36 (ans).
We have 1.00 L of the buffer solution; therefore,
Moles HCO3- = (volume of buffer in L)*(molarity of HCO3-) = (1.00 L)*(0.360 M) = 0.360 mole
Moles CO32- = (volume of buffer in L)*(molarity of CO32-) = (1.00 L)*(0.397 M) = 0.397 mole.
HNO3 ionizes completely to furnish H+ which reacts with NH3 as below.
HNO3 (aq) --------> H+ (aq) + NO3- (aq)
H+ (aq) + CO32- (aq) --------> HCO3- (aq)
As per the stoichiometric equation,
moles H+ = moles HNO3 = moles CO32- neutralized = moles HCO3-formed = 0.108 mole.
Moles CO32- after the addition of HNO3 = (0.397 – 0.108) mole = 0.289 mole.
Moles HCO3- formed after the addition = (0.360 + 0.108) mole = 0.468 mole.
Since the volume of the buffer is 1.00 L, therefore,
[CO32-] = (moles of CO32-)/(volume of the buffer) = (0.289 mole)/(1.00 L) = 0.289 M.
[HCO3-] = (moles of HCO3-)/(volume of the buffer) = (0.468 mole)/(1.00 L) = 0.468 M.
Use the Henderson-Hasslebach equation to find the pH.
pH = pKa + log [CO32-]/[HCO3-]
= 10.32 + log (0.289 M)/(0.468 M)
= 10.32 + log (0.6175)
= 10.32 + (-0.2094)
= 10.32 – 0.2094 = 10.1106 ? 10.11 (ans).
Change in pH of the buffer after the addition of HNO3 = (new pH) – (old pH) = (10.11) – (10.36) = -0.25 (ans).
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