Question
In: Chemistry
what is the ph when 25.00 no of .20 M CH3COOH has been treated with 40.0...
what is the ph when 25.00 no of .20 M CH3COOH has been treated with 40.0 no of .10 M NaOH?
Solutions
Expert Solution
no of moles of CH3COOH = molarity * volume in L
= 0.2*0.025 = 0.005 moles
no of moles of NaOH = molarity * volume in L
= 0.1*0.04 = 0.004 moles
CH3COOH + NaOH --------------------> CH3COONa + H2O
I 0.005 0.004 0
C -0.004 -0.004 0.004
E 0.001 0 0.004
[CH3COONa] = 0.004 moles
[CH3COOH] = 0.001 moles
PH = Pka + log[CH3COONa]/[CH3COOH]
= 4.75 + log0.004/0.001
= 4.75 + 0.6020 = 5.352 >>>>>>answer
queen_honey_blossom answered 2 years agoRelated Solutions
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