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Q. Fixed volume of 0.1 M benzoic acid (pKa = 4.2) solution is added into 0.2 M sodium benzoate solution and formed a 300 mL. resulting acidic buffer solution. If pH of the resulting solution is 3.9, then added volume of benzoic acid is 

   (a) 240 mL                                                             (b) 150 mL

   (c) 100 mL                                                             (d) None‚Äč
 
 

Ph = pka + log salt/acid 3.9 = 4.2 +log salt/acid Log salt/acid = -.3 Salt/acid =.5 .5 =300-V/V From here we find the value V If any problem contact me 7986558520
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150ml
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Answer = 100 mL pH=pKa+log[Salt][Acid] Putting the given values, 4.5=4.2+log[Salt][Acid] 0.3=log[Salt][Acid] [Salt][Acid] = antilog 0.3 = 2 Let v mL of benzoic acid is mixed with (300-v) mL of 1M sodium benzoate. ∴300−vv=2 ⇒300−v=2v v = 100 mL
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