Question

for the first question use 0.399 , for the second question use 0.0399

1. What is the pH of the solution which results from mixing 50.0 mL of 0.XXX M HF(aq) and 50.0 mL of 0.XXX M NaOH(aq) at 25 °C? (K, of HF = 7.2 x 10 ) (XXX is your last three digit in your T number. For example, if your last 3 digit is 763, then 0.xXX-0.763. You need to show your calculation to get full points) 2. What mass of solid NACHCO2 (molar mass = 82.0 g/mol) should be added to 1.0 L of 0.0XXX M CH CO2H to make a buffer with a pH of 7.98? (pkK, of CH CO2H = 7.21) (XXX is your last three digit in your T number. For example, if your last 3 digit is 763, then 0.0XXX-0.0763. You need to show your calculation to get full points).

Answer 1

1. Ans :-

Number of moles of HF = Molarity x Volume of solution in L

= 0.399 M x 0.050 L

= 0.01995 mol

and

Number of moles of NaOH = Molarity x Volume of solution in L

= 0.399 M x 0.050 L

= 0.01995 mol

ICF table :

..........................HF..............+..............NaOH -------------------------> NaF..................+................H₂O

Initial ................0.01995 mol...............0.01995 mol...........................0.0 mol.............................

Change............-0.01995 mol..............-0.01995 mol............................+0.01995 mol.................

Final......................0.0 mol......................0.0 mol..................................0.01995 mol

Moles of NaF formed = 0.01995 mol

Molarity of NaF or F^-1= 0.01995 mol / 0.100 L

= 0.1995 M

ICE table :

..........................F^-..............+..............H₂O ----------------------------> HF..................+................OH^-

Initial ................0.1995 M.............................................................0.0 mol.....................................0.0 M

Change............-y.............................................................................+y..........................................+y

Final................(0.1995-y) M..............................................................y M.......................................y M

y = Amount dissociated per mole

Expression of acid dissociation constant K_a is :

K_a = [OH^-] .[HF] / [F^-]

7.2 x 10^-4 = y² / (0.1995-y)

y² + 7.2 x 10^-4 y - 1.4364 x 10^-4 = 0

On solving

y = 0.01163

So, [OH^-] = 0.01163 M

pOH = - log [OH^-]

= - log 0.01163 M

= 1.93

So,

pH = 14 - pOH

pH = 14 - 1.93

pH = 12.07

So,

pH = 12.07  

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2. Sol :-

Given, Molarity of CH₃CO₂H = 0.0399 M

Using Henderson-Hasselbalch equation :

pH = pK_a + log [CH₃CO₂Na] / [CH₃CO₂​​​​​​​H]

7.98 = 7.21 + log [CH₃CO₂​​​​​​​Na] / 0.0399 M

log [CH₃CO₂​​​​​​​Na] / 0.0399 M = 7.98 - 7.21

log [CH₃CO₂​​​​​​​Na] - log 0.0399 M = 0.77

log [CH₃CO₂​​​​​​​Na] + 1.40  = 0.77

log [CH₃CO₂​​​​​​​Na] = 0.77 - 1.40

log [CH₃CO₂​​​​​​​Na] = - 0.63

[CH₃CO₂​​​​​​​Na] = 10^-0.63 M

[CH₃CO₂​​​​​​​Na] = 0.234 M

Moles of CH₃CO₂​​​​​​​Na = Molarity x Volume in L

= 0.234 M x 1.00 L

= 0.234 mol

So,

Mass of CH₃CO₂​​​​​​​Na = Moles x Gram molar mass

= 0.234 mol x 82.0 g/mol

= 19.2 g

So,

Mass of CH₃CO₂​​​​​​​Na added to the buffer = 19.2 g

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