1. Buffer Problem... calculate the pH of the buffer and then the pH after adding
ID: 997950 • Letter: 1
Question
1. Buffer Problem... calculate the pH of the buffer and then the pH after adding a base to 50.0 mL the buffer. Which pair of solutions can be mixed together to form a buffer? Circle the pairs that can be mixed together to form a buffer solution? Nacl/HCI KOH/ HF HCI/NHs CSF/H NaOH/NaC2H,O2 Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Ka of benzoic acid = 6.3 × 10-5. .Calculate the pH of the buffer after the addition of 13.0 mL of 0.15 M HCI to 50.0 mL of a buffer. When calculating concentration, be sure to use the total volume of 63.0 mL. Calculate the pH of the buffer after the addition of 13.0 mL of 0.15 M KOH to 50.0 mL of a buffer. When calculating concentration, be sure to use the total volume of 63.0 mL.Explanation / Answer
(1)(i)
Buffer is a mixture of weak acid and its conjugate base or a weak base and its conjugate acid.
HCl is a strong acid and Cl- is its conjugate base so, NaCl/HCl is not a good pair to form a buffer solution.
HF is a weak acid and KOH is a strong base. So, KOH/HF pair could form a buffer since KOH react with HF to form KF and the mixture of KF and HF would work as a buffer solution.
HCl/NH3 pair would also form a buffer solution since HCl is a strong acid and NH3 is a weak base. on their reaction we get NH4Cl and the mixture of NH4Cl and NH3 acts as a buffer solution.
CaF/HF would also act as a buffer solution since HF is a weak acid and F- is its conjugate base (CaF is a salt).
NaOH is a strong base and NaC2H3O2 is also a base so NaOH/NaC2H3O2 pair would not form a buffer solution.
(ii) we use Handerson equation to calculate the pH of a buffer solution..
pH = Pka + log(base/acid)
Pka = - log(ka) = - log(6.3 x 10-5) = 4.2
pH = 4.2 + log(0.20/0.12)
pH = 4.2 + 0.22 = 4.42
(iii)
initial moles of benzoic acid = 0.12 x 0.050 = 0.006
initial moles of sodium benzoate or benzoate ion = 0.20 x 0.050 = 0.010
moles of HCl added = 0.013 x 0.15 = 0.00195
HCl is a strong acid so it would give H+ and this would react with benzoate ion to form benzoic acid and reaction takes place in 1:1 ratio.
So, new moles of benzoic acid = 0.006 + 0.00195 = 0.00795
new moles of benzoate ion = 0.010 - 0.00195 = 0.00805
concentration of benzoic acid = 0.00795/0.063 = 0.126
concentration of benzoate ion = 0.128 M
pH = 4.2 + log(0.128/0.126)
pH = 4.2 + 0.0068 = 4.2068
(iv)
initial moles of benzoic acid = 0.12 x 0.050 = 0.006
initial moles of sodium benzoate or benzoate ion = 0.20 x 0.050 = 0.010
moles of KOH added = 0.013 x 0.15 = 0.00195
KOH is a strong base so it would give OH- and this would react with benzoic acid to form benzoate ion and reaction takes place in 1:1 ratio.
So, new moles of benzoate ion = 0.006 + 0.00195 = 0.00795
new moles of benzoic acid = 0.010 - 0.00195 = 0.00805
concentration of benzoate ion = 0.00795/0.063 = 0.126
concentration of benzoic acid = 0.00805/0.063 = 0.128 M
pH = 4.2 + log(0.126/0.128)
pH = 4.2 - 0.0068 = 4.1932
(2) (i)
moles of HCl = 0.025 x 0.100 = 0.00250
moles of KOH added = 0.0158 x 0.150 = 0.00237
KOH and HCl react in 1:1 ratio.
since KOH is limiting. so, 0.00237 moles of it would neutralize exactly 0.00237 moles of HCl.
excess moles of HCl = 0.00250 - 0.00237 = 0.00013
total volume = 0.025 L + 0.0158 L = 0.0408
concentration of HCl = 0.00013/0.0408 = 0.00319M
pH = - log(0.00319) = 2.496 or it could be round to 2.50
(ii)
Initially we have 0.00250
at equivalence point we have equal moles of acid and base.
so, moles of KOH would also be 0.00250
molarity = mole/L
L = moles/molarty
So, 0.00250/0.150 = 0.0167 L = 16.7 ml
(3) (i)
The solubilities of those salts would increase in acidic solution that has conjugate base of a weak acid.
ZnCO3 ------------> Zn+2 + CO3-2 (this is the conjugate base of weak acid so solubility will increase)
ZnS -----------> Zn+2 + S-2 (this is also the conjugate base of weak acid so solubility will increase)
AgCl ---------> Ag+ + Cl- (this is the conjugate base of strong acid so solubility will not increase)
Ba3(PO4)2 -------->3 Ba+2 + 2PO4-3(conjugate base of weak acid so solubility will increase)
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