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50 ml 0.1 M NaOH is added to 50 ml of 0.1 M CH3COOH solution, what will be the final pH? please give me a detailed solution for this given that pKa=pKb= 4.7447 pKw= 14

50 ml 0.1 M NaOH is added to 50 ml of 0.1 M CH3COOH solution, what will be the final pH? please give me a detailed solution for this given that pKa=pKb= 4.7447 pKw= 14

Grade:11

3 Answers

Mehruna Bipasha
33 Points
10 years ago
  • after adding 50ml 0.1M NaOH in 50ml of 0.1M CH3COOH solution,there willbe made 50ml 0.1 M CH3COONa & there remaining 50ml 0.1M CH3COOH. total velocity=(50+50+50)=150ml.Molarity of CH3COOH=(50x0.1)/150=.0333M.Molarity of CH3COONa=(50x0.1)/150=.0333M.then pH=pKa+log [salt]/[acid]=4.7447+log [.0333]/[.0333]=4.7447+log 1=4.7447+0=4.7447.So the ans is pH=4.7447.

Pawan
21 Points
3 years ago
NaOH reacts with CH3COOH in 1:1 molar ratio. You are mixing equimolar amounts of the two reactants . You will produce a solution of CH3COONa Mol CH3COOH in 50.0mL of 0.1M solution = 50/1000*0.1 = 0.005 mol CH3COOH This will produce 0.005 mol CH3COONa dissolved in 100mL = 0.1L solution Molarity of CH3COONa solution = 0.005/0.1 = 0.05M CH3COONa solution. Ka CH3COOH = 1.8*10^-5 Kb = 10^-14 / ( 1.8*10^-5) Kb = 5.56*10^-10 Use Kb equation to calculate [OH-] Kb = [OH-]² / [CH3COONa] 5.56*10^-10 = [OH-]² / 0.05 [OH-]² = (5.56*10^-10)* 0.05 [OH-]² = 2.78*10^-11 [OH-] = 5.27*10^-6 pOH = - log [OH-] pOH = -log ( 5.27*10^-6) pOH = 5.28 pH = 14 - pOH pH = 14 - 5.28 pH = 8.72...this will be wright ans
ankit singh
askIITians Faculty 614 Points
one year ago

Answer

NaOH+CH3COOHCH3COONa+H2O

no. of mili equivalents = molarity × volume × n-factor
Meq. of NaOH=0.1×60×1=6
Meq. of CH3COOH=0.1×50×1=5
          NaOH+CH3COOHCH3COONa+H2O
initial       6                5 
final         1                0                           5                       5
In solution, we have strong base. so salt CH3COONa is neglected.
(SO) BOH=log[OH]
no. of mili equivalent of NaOH=1
so, 1= normality × volume
      1=N×110
   1101=N  
pOH=log[1/110]=2.04
So, pH=11.96

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