50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10-2. (Nearest integer)
(Given : pKa (CH3-COOH) = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04
50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10-2. (Nearest integer)
(Given : pKa (CH3-COOH) = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04
Here, total meq of acetic acid = 50 * 0.1 = 5
And total meq of NaOH = 25 * 0.1 = 2.5
After neutralization process
Meq of left acetic acid = 2.5
And meq of formed CH3COONa = 2.5
Similar Questions for you
0.01 M NaOH,
M = 1 * 10-2

pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
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Chemistry Ncert Solutions Class 11th 2023
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