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Q128  Ans1
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Asked by manasvijha | 24 Apr, 2019, 09:23: AM
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Option (1) is correct.
 
Given:
 
Volume of NaOH = 20 ml
 
 
Molaity of NaOH = 0.1 M
 
Volume of CH3COOH = 30 ml
 
Molarity of CH3COOH = 0.2 M
 
CH3COOH + NaOH → CH3COONa + H2O
 
 
20 ml of 0.1 M NaOH contains 2 milli moles of NaOH
 
30 ml of 0.2 M CH3COOH contains 6 milli mole of CH3COOH
 
CH3COOH left unreacted = 6-2 = 4 millimole
 
CH3COONa formed = 2 milli mole
 
Volume of solution = 50 ml
 
left square bracket CH subscript 3 COOH right square bracket space equals space 4 over 50 space equals space 2 over 25 space straight M left square bracket CH subscript 3 COONa right square bracket space equals space 2 over 50 space space equals space 1 over 25 straight M
 
 
pH space equals space pK subscript straight a space plus space log fraction numerator open square brackets Salt close square brackets over denominator open square brackets Acid close square brackets end fraction space space space space space space equals space 4.74 space plus space log fraction numerator begin display style bevelled 1 over 25 end style over denominator begin display style bevelled 2 over 25 end style end fraction space space space space space space space equals space 4.74 space plus log space open parentheses 0.5 close parentheses space space space space space space space equals space 4.74 minus 0.301 space pH space equals 4.44
 
The pH of the resulting solution will be 4.44
Answered by Varsha | 24 Apr, 2019, 02:36: PM

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