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40 ml 0.1 M Ba ( HCOO) is mixed with 20ml 0.1M HCOOH. Find the pH of the mixture. Given Ka of HCOOH = 10-5 at 25°C.

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We have given,

40 ml 0.1 M Ba (HCOO)2

20 Ml 0.1 M HCOOH

Ka value of HCOOH = 10-5

Numbers of moles of  HCOO- = 2 x 40 ml  x 10-3 x 0.1 mole

= 8 millimole

Number of moles of HCOOH = 20 x 100-3 x 0.1 mole

= 2 millimole

After mixing, the volume of solution = (40 + 20) ml

= 60 ml

the concentration of HCOO- (salt) = \(\cfrac{8}{60}\)

= 0.133 M

the concentration of HCOOH (acid) = \(\cfrac2{60}\)

= 0.0 33 M

pKa at HCOOH = - log Ka

- log 10-5

= 5

Using Henderson  Hesselbalch equation 

pH = pKa + log \(\cfrac{[salt]}{[acid]}\) 

5 + log \(\cfrac{0.133}{0.033}\)

= 5.6

Hence pH of solution will be 5.6

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